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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received n resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate. Input The only line of the input contains one integer n (7 ≤ n ≤ 777) — the number of potential employees that sent resumes. Output Output one integer — the number of different variants of group composition. Examples Input 7 Output 29 Submitted Solution: ``` n = int(input()) fac_n = 1 fac1 = 1 fac2 = 1 fac3 = 1 i = 0 while i < n: i += 1 fac_n = fac_n * i i = 0 while i < n - 5: i += 1 fac1 = fac1 * i i = 0 while i < n - 6: i += 1 fac2 = fac2 * i i = 0 while i < n - 7: i += 1 fac3 = fac3 * i ans = fac_n / (fac1 * 120)+ fac_n / (fac2 * 720) + fac_n / (fac3 * 5040) print(ans) ```
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received n resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate. Input The only line of the input contains one integer n (7 ≤ n ≤ 777) — the number of potential employees that sent resumes. Output Output one integer — the number of different variants of group composition. Examples Input 7 Output 29 Submitted Solution: ``` n = int(input()) print(n * (n - 1) * (n - 2) * (n - 3) * (n - 4) / 1 / 2 / 3 / 4 / 5 + n * (n - 1) * (n - 2) * (n - 3) * (n - 4) * (n - 5) / 1 / 2 / 3 / 4 / 5 / 6 + n * (n - 1) * (n - 2) * (n - 3) * (n - 4) * (n - 5) * (n - 6) / 1 / 2 / 3 / 4 / 5 / 6 / 7) ```
instruction
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83,008
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No
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1
83,008
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166,017
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received n resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate. Input The only line of the input contains one integer n (7 ≤ n ≤ 777) — the number of potential employees that sent resumes. Output Output one integer — the number of different variants of group composition. Examples Input 7 Output 29 Submitted Solution: ``` a=int(input()) pr=a*(a-1)*(a-2)*(a-3)*(a-4) print(int(pr/120+pr*(a-5)/720+pr*(a-5)*(a-6)/5040)) ```
instruction
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83,009
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No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` a = [int(i) for i in input().split()] a1 = [[a[(j * 4) + i] for i in range(4)] for j in range(6)] ver = a1[0] per = a1[1] niz = a1[2] lev = a1[3] prav = a1[4] zad = a1[5] res = "NO" if len(set(lev)) == 1 and len(set(prav)) == 1: per1 = (1 if per[1] == per[3] else 0) ver1 = (1 if ver[1] == ver[3] else 0) niz1 = (1 if niz[1] == niz[3] else 0) zad1 = (1 if zad[0] == zad[2] else 0) per2 = (1 if per[0] == per[2] else 0) ver2 = (1 if ver[0] == ver[2] else 0) niz2 = (1 if niz[0] == niz[2] else 0) zad2 = (1 if zad[1] == zad[3] else 0) if (per1 == ver2 and per[1] == ver[0]) and (ver1 == zad2 and zad[1] == ver[1]) and (zad1 == niz2 and zad[0] == niz[0]) and (niz1 == per2 and per[0] == niz[1]): res = "YES" elif (per1 == niz2 and per[1] == niz[0]) and (ver1 == per2 and per[0] == ver[1]) and (zad1 == ver2 and zad[0] == ver[0]) and (niz1 == zad2 and zad[1] == niz[1]): res = "YES" elif len(set(niz)) == 1 and len(set(ver)) == 1: per1 = (1 if per[0] == per[1] else 0) lev1 = (1 if lev[0] == lev[1] else 0) prav1 = (1 if prav[0] == prav[1] else 0) zad1 = (1 if zad[0] == zad[1] else 0) per2 = (1 if per[2] == per[3] else 0) lev2 = (1 if lev[2] == lev[3] else 0) prav2 = (1 if prav[2] == prav[3] else 0) zad2 = (1 if zad[2] == zad[3] else 0) if (per1 == lev2 and per[0] == lev[2]) and (lev1 == zad2 and zad[2] == lev[0]) and (zad1 == prav2 and zad[0] == prav[2]) and (prav1 == per2 and per[2] == prav[0]): res = "YES" elif (per1 == prav2 and per[0] == prav[2]) and (prav1 == zad2 and zad[2] == prav[0]) and (zad1 == lev2 and zad[0] == lev[2]) and (lev1 == per2 and per[2] == lev[0]): res = "YES" elif len(set(per)) == 1 and len(set(zad)) == 1: ver1 = (1 if ver[2] == ver[3] else 0) lev1 = (1 if lev[3] == lev[1] else 0) prav1 = (1 if prav[0] == prav[2] else 0) niz1 = (1 if niz[0] == niz[1] else 0) ver2 = (1 if ver[0] == ver[1] else 0) lev2 = (1 if lev[2] == lev[0] else 0) prav2 = (1 if prav[1] == prav[3] else 0) niz2 = (1 if niz[2] == niz[3] else 0) if (ver1 == lev2 and ver[2] == lev[2]) and (lev1 == niz2 and niz[2] == lev[3]) and (niz1 == prav2 and niz[0] == prav[1]) and (prav1 == ver2 and ver[0] == prav[0]): res = "YES" elif (ver1 == prav2 and ver[2] == prav[1]) and (prav1 == niz2 and niz[2] == prav[0]) and (niz1 == lev2 and niz[0] == lev[2]) and (lev1 == ver2 and ver[0] == lev[3]): res = "YES" if len(set(per)) == 1 and len(set(zad)) == 1 and len(set(niz)) == 1 and len(set(ver)) == 1 and len(set(lev)) == 1 and len(set(prav)) == 1: res = "NO" print(res) ```
instruction
0
83,099
11
166,198
Yes
output
1
83,099
11
166,199
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` a = list(map(int, input().split())) if ((a[0] == a[5] == a[2] == a[7]) and (a[4] == a[9] == a[6] == a[11]) and (a[20] == a[8] == a[10] == a[22]) and (a[21] == a[1] == a[23] == a[3])) or \ ((a[4] == a[1] == a[6] == a[3]) and (a[8] == a[5] == a[10] == a[7]) and (a[21] == a[23] == a[9] == a[11]) and (a[0] == a[20] == a[2] == a[22])) or \ ((a[4] == a[5] == a[14] == a[15]) and (a[16] == a[6] == a[17] == a[7]) and (a[20] == a[18] == a[21] == a[19]) and (a[22] == a[12] == a[13] == a[23])) or\ ((a[12] == a[13] == a[6] == a[7]) and (a[4] == a[5] == a[19] == a[18]) and (a[16] == a[17] == a[22] == a[23]) and (a[15] == a[14] == a[21] == a[20])) or \ ((a[2] == a[3] == a[17] == a[19]) and (a[11] == a[18] == a[10] == a[16]) and (a[8] == a[12] == a[9] == a[14]) and (a[1] == a[15] == a[0] == a[13])) or \ ((a[16] == a[0] == a[1] == a[18]) and (a[3] == a[2] == a[12] == a[14]) and (a[11] == a[15] == a[10] == a[13]) and (a[8] == a[17] == a[9] == a[19])): print('YES') else: print("NO") ```
instruction
0
83,100
11
166,200
Yes
output
1
83,100
11
166,201
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` def checkRubik(l): for i in range(1,25,4): for j in range(i+1,i+4): if l[j]!=l[i]: return False return True a=list(map(int,input().split())) a.insert(0,0) o=checkRubik(a) if not o: b=a b[1],b[3],b[5],b[7],b[9],b[11],b[24],b[22]=b[24],b[22],b[1],b[3],b[5],b[7],b[9],b[11] o=checkRubik(b) if not o: b=a b[1],b[3],b[5],b[7],b[9],b[11],b[24],b[22]=b[5],b[7],b[9],b[11],b[24],b[22],b[1],b[3] o=checkRubik(b) if not o: b=a b[2],b[4],b[6],b[8],b[10],b[12],b[23],b[21]=b[23],b[21],b[2],b[4],b[6],b[8],b[10],b[12] o=checkRubik(b) if not o: b=a b[2],b[4],b[6],b[8],b[10],b[12],b[23],b[21]=b[6],b[8],b[10],b[12],b[23],b[21],b[2],b[4] o=checkRubik(b) if not o: b=a b[13],b[14],b[5],b[6],b[17],b[18],b[21],b[22]=b[21],b[22],b[13],b[14],b[5],b[6],b[17],b[18] o=checkRubik(b) if not o: b=a b[13],b[14],b[5],b[6],b[17],b[18],b[21],b[22]=b[5],b[6],b[17],b[18],b[21],b[22],b[13],b[14] o=checkRubik(b) if not o: b=a b[15],b[16],b[7],b[8],b[19],b[20],b[23],b[24]=b[23],b[24],b[15],b[16],b[7],b[8],b[19],b[20] o=checkRubik(b) if not o: b=a b[15],b[16],b[7],b[8],b[19],b[20],b[23],b[24]=b[7],b[8],b[19],b[20],b[23],b[24],b[15],b[16] o=checkRubik(b) #start if not o: b=a b[3],b[4],b[17],b[19],b[10],b[9],b[16],b[14]=b[16],b[14],b[3],b[4],b[17],b[19],b[10],b[9] o=checkRubik(b) if not o: b=a b[3],b[4],b[17],b[19],b[10],b[9],b[16],b[14]=b[17],b[19],b[10],b[9],b[16],b[14],b[3],b[4] o=checkRubik(b) if not o: b=a b[1],b[2],b[18],b[20],b[12],b[11],b[15],b[13]=b[15],b[13],b[1],b[2],b[18],b[20],b[12],b[11] o=checkRubik(b) if not o: b=a b[1],b[2],b[18],b[20],b[12],b[11],b[15],b[13]=b[18],b[20],b[12],b[11],b[15],b[13],b[1],b[2] o=checkRubik(b) if not o: print("NO") else: print("YES") else: print("NO") ```
instruction
0
83,101
11
166,202
Yes
output
1
83,101
11
166,203
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` l=list(map(int,input().split())) l.insert(0,0) c1=[1,6,3,8,5,10,7,12,9,23,11,21,13,14,15,16,17,18,19,20,4,22,2,24] c2=[1,23,3,21,5,2,7,4,9,6,11,8,13,14,15,16,17,18,19,20,12,22,10,24] c3=[1,2,3,4,5,6,15,16,9,10,11,12,13,14,23,24,17,18,7,8,21,22,19,20] c4=[1,2,3,4,5,6,19,20,9,10,11,12,13,14,7,8,17,18,23,24,21,22,15,16] c5=[1,2,16,14,5,6,7,8,19,17,11,12,13,9,15,10,3,18,4,20,21,22,23,24] c6=[1,2,17,19,5,6,7,8,14,16,11,12,13,4,15,3,10,18,9,20,21,22,23,24] flag=0 mark=0 for i in range(6): if(l[c1[4*i]] == l[c1[4*i+1]] == l[c1[4*i+2]] == l[c1[4*i+3]]): mark=1 else: mark=0 break if(mark): flag=1 mark=0 for i in range(6): if(l[c2[4*i]] == l[c2[4*i+1]] == l[c2[4*i+2]] == l[c2[4*i+3]]): mark=1 else: mark=0 break if(mark): flag=1 mark=0 for i in range(6): if(l[c3[4*i]] == l[c3[4*i+1]] == l[c3[4*i+2]] == l[c3[4*i+3]]): mark=1 else: mark=0 break if(mark): flag=1 mark=0 for i in range(6): if(l[c4[4*i]] == l[c4[4*i+1]] == l[c4[4*i+2]] == l[c4[4*i+3]]): mark=1 else: mark=0 break if(mark): flag=1 mark=0 for i in range(6): if(l[c5[4*i]] == l[c5[4*i+1]] == l[c5[4*i+2]] == l[c5[4*i+3]]): mark=1 else: mark=0 break if(mark): flag=1 mark=0 for i in range(6): if(l[c6[4*i]] == l[c6[4*i+1]] == l[c6[4*i+2]] == l[c6[4*i+3]]): mark=1 else: mark=0 break if(mark): flag=1 if(flag): print("YES") else: print("NO") ```
instruction
0
83,102
11
166,204
Yes
output
1
83,102
11
166,205
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` def f(a, d): newa = a[::] for i in range(24): if (i + 1) in d: newa[d[i + 1] - 1] = a[i] #print(newa) for i in range(0, 22, 4): if newa[i] == newa[i + 1] == newa[i + 2] == newa[i + 3]: return True return False arr = list(map(int, input().split())) var = [dict() for i in range(8)] #1 var[0][2] = 6 var[0][4] = 8 var[0][6] = 10 var[0][8] = 12 var[0][10] = 23 var[0][12] = 21 var[0][23] = 2 var[0][21] = 4 var[0][17] = 19 var[0][19] = 20 var[0][20] = 18 var[0][18] = 17 #3 var[2][1] = 5 var[2][3] = 7 var[2][5] = 9 var[2][7] = 11 var[2][9] = 24 var[2][11] = 22 var[2][24] = 1 var[2][22] = 3 var[2][13] = 14 var[2][15] = 13 var[2][16] = 15 var[2][14] = 16 #5 var[4][13] = 5 var[4][14] = 6 var[4][5] = 17 var[4][6] = 18 var[4][17] = 21 var[4][18] = 22 var[4][21] = 13 var[4][22] = 14 var[4][1] = 3 var[4][3] = 4 var[4][4] = 2 var[4][2] = 1 #7 var[6][15] = 7 var[6][16] = 8 var[6][7] = 19 var[6][8] = 20 var[6][19] = 23 var[6][20] = 24 var[6][23] = 15 var[6][24] = 16 var[6][9] = 10 var[6][10] = 12 var[6][12] = 11 var[6][11] = 9 for i in range(0, 8, 2): for j in range(1, 25): if j in var[i]: var[i + 1][var[i][j]] = j for i in range(8): #print(i - 1) if f(arr, var[i]): print('YES') exit() print('NO') ```
instruction
0
83,103
11
166,206
No
output
1
83,103
11
166,207
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` # by the authority of GOD author: Kritarth Sharma # import math,copy from collections import defaultdict,Counter from itertools import permutations def fact(n): return 1 if (n == 1 or n == 0) else n * fact(n - 1) def prime(n) : if (n <= 1) : return False if (n <= 3) : return True if (n % 2 == 0 or n % 3 == 0) : return False i = 5 while(i * i <= n) : if (n % i == 0 or n % (i + 2) == 0) : return False i = i + 6 return True def inp(): l=list(map(int,input().split())) return l def one(l,i,s): a=False b=False for ii in range(i-1,-1,-1): if l[ii]==0 and s[ii]==0: a=ii break for ii in range(i+1,len(l)): if l[ii]==0 and s[ii]==0: b=ii break if a==False: return b elif b==False: return a elif abs(a-i)>abs(b-i): return b else: return a ## code starts here ################################################################################################################# def main(): l=inp() p=[0]*25 for i in range(1,25): p[i]=l[i-1] if p[17]==p[18]==p[19]==p[20] and p[13]==p[14]==p[15]==p[16] and p[1]==p[2]==p[3]==p[4] and p[9]==p[10]==p[11]==p[12] and p[5]==p[6]==p[7]==p[8] and p[21]==p[22]==p[23]==p[24]: print("YES") elif p[17]==p[18]==p[19]==p[20] and p[13]==p[14]==p[15]==p[16]: if p[5]==p[7]==p[10]==p[12] and p[9]==p[11]==p[21]==p[23] and p[2]==p[4]==p[22]==p[24] and p[1]==p[3]==p[6]==p[8]: print("YES") elif p[5]==p[7]==p[2]==p[4] and p[9]==p[11]==p[6]==p[8] and p[12]==p[10]==p[22]==p[24] and p[1]==p[3]==p[21]==p[23]: print("YES") else: print("NO") elif p[1]==p[2]==p[3]==p[4] and p[9]==p[10]==p[11]==p[12]: if p[5]==p[6]==p[19]==p[20] and p[7]==p[8]==p[13]==p[14] and p[17]==p[18]==p[23]==p[24] and p[21]==p[22]==p[16]==p[15]: print("YES") elif p[5]==p[6]==p[15]==p[16] and p[17]==p[18]==p[7]==p[8] and p[19]==p[20]==p[22]==p[21] and p[14]==p[13]==p[24]==p[23]: print("YES") else: print("NO") elif p[5]==p[6]==p[7]==p[8] and p[21]==p[22]==p[23]==p[24]: if p[19]==p[17]==p[1]==p[2] and p[14]==p[16]==p[18]==p[20] and p[13]==p[15]==p[9]==p[10] and p[11]==p[12]==p[3]==p[4]: print("YES") elif p[18]==p[20]==p[3]==p[4] and p[17]==p[19]==p[13]==p[15] and p[11]==p[12]==p[14]==p[16] and p[1]==p[2]==p[9]==p[10]: print("YES") else: print("NO") else: print("NO") import os,sys from io import BytesIO,IOBase #Fast IO Region BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") if __name__ == "__main__": main() def random(): """My code gets caught in plagiarism check for no reason due to the fast IO template, . Due to this reason, I am making useless functions""" rating=100 rating=rating*100 rating=rating*100 print(rating) def random(): """My code gets caught in plagiarism check for no reason due to the fast IO template, . Due to this reason, I am making useless functions""" rating=100 rating=rating*100 rating=rating*100 print(rating) def random(): """My code gets caught in plagiarism check for no reason due to the fast IO template, . Due to this reason, I am making useless functions""" rating=100 rating=rating*100 rating=rating*100 print(rating) def random(): """My code gets caught in plagiarism check for no reason due to the fast IO template, . Due to this reason, I am making useless functions""" rating=100 rating=rating*100 rating=rating*100 print(rating) ```
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No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. During the breaks between competitions, top-model Izabella tries to develop herself and not to be bored. For example, now she tries to solve Rubik's cube 2x2x2. It's too hard to learn to solve Rubik's cube instantly, so she learns to understand if it's possible to solve the cube in some state using 90-degrees rotation of one face of the cube in any direction. To check her answers she wants to use a program which will for some state of cube tell if it's possible to solve it using one rotation, described above. Cube is called solved if for each face of cube all squares on it has the same color. https://en.wikipedia.org/wiki/Rubik's_Cube Input In first line given a sequence of 24 integers ai (1 ≤ ai ≤ 6), where ai denotes color of i-th square. There are exactly 4 occurrences of all colors in this sequence. Output Print «YES» (without quotes) if it's possible to solve cube using one rotation and «NO» (without quotes) otherwise. Examples Input 2 5 4 6 1 3 6 2 5 5 1 2 3 5 3 1 1 2 4 6 6 4 3 4 Output NO Input 5 3 5 3 2 5 2 5 6 2 6 2 4 4 4 4 1 1 1 1 6 3 6 3 Output YES Note In first test case cube looks like this: <image> In second test case cube looks like this: <image> It's possible to solve cube by rotating face with squares with numbers 13, 14, 15, 16. Submitted Solution: ``` def Pedy(arr1, arr2): print(arr1) print(arr2) for i in range(8): h = i if i % 2: continue hh = 0 anss = 1 while (h != ((i - 1) + 8) % 8 and anss): if arr1[h] != arr2[hh]: anss = 0 h += 1 hh += 1 if h == 8: h = 0 if anss and arr1[h] == arr2[hh] and (i == 0 or i == 2 or i == 6 or i == 4): return 1 return 0 a = list(map(int, input().split())) b = [] for j in range(6): i = j * 4 if a[i] == a[i + 1] and a[i + 1] == a[i + 2] and a[i + 2] == a[i + 3]: b.append(0) else: b.append(1) c = [] c.append(b[0] or b[2]) c.append(b[1] or b[5]) c.append(b[3] or b[4]) if c[0] + c[1] + c[2] == 3 or c[0] + c[1] + c[2] == 1: print("NO") if c[0] + c[1] + c[2] == 0: print("YES") if c[0] + c[1] + c[2] == 2: if c[0] + c[1] == 2: arr1 = [] arr2 = [] arr1.append(a[0]) arr1.append(a[2]) arr1.append(a[4]) arr1.append(a[6]) arr1.append(a[8]) arr1.append(a[10]) arr1.append(a[21]) arr1.append(a[23]) arr2.append(a[1]) arr2.append(a[3]) arr2.append(a[5]) arr2.append(a[7]) arr2.append(a[9]) arr2.append(a[11]) arr2.append(a[20]) arr2.append(a[22]) if Pedy(arr1, arr2): print("YES") else: print("NO") if c[0] + c[2] == 2: arr1 = [] arr2 = [] arr1.append(a[0]) arr1.append(a[1]) arr1.append(a[17]) arr1.append(a[19]) arr1.append(a[11]) arr1.append(a[10]) arr1.append(a[14]) arr1.append(a[12]) arr2.append(a[2]) arr2.append(a[3]) arr2.append(a[16]) arr2.append(a[18]) arr2.append(a[9]) arr2.append(a[8]) arr2.append(a[15]) arr2.append(a[13]) if Pedy(arr1, arr2): print("YESs") else: print("NOo") if c[1] + c[2] == 2: arr1 = [] arr2 = [] arr1.append(a[12]) arr1.append(a[13]) arr1.append(a[4]) arr1.append(a[5]) arr1.append(a[16]) arr1.append(a[17]) arr1.append(a[20]) arr1.append(a[21]) arr2.append(a[14]) arr2.append(a[15]) arr2.append(a[6]) arr2.append(a[7]) arr2.append(a[18]) arr2.append(a[19]) arr2.append(a[22]) arr2.append(a[23]) if Pedy(arr1, arr2): print("YES") else: print("NO") ```
instruction
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No
output
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166,211
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem You have to play a darts game with the following rules. You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points. Create a program that finds the maximum number of points you can get given the written points and the value of M. input The input consists of multiple datasets. Each dataset is given in the following format. On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written. Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300. When both N and M are 0, the input is completed. The number of data sets does not exceed 10. output Outputs the maximum number of points you can get for each dataset on one line. Examples Input 4 50 3 14 15 9 3 21 16 11 2 0 0 Output 48 20 Input None Output None Submitted Solution: ``` # http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0529 # http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=2985387#1 import bisect while True: n, m = map(int, input().split()) if n == 0: break p = [int(input()) for i in range(n)] p.append(0) p.sort() s = [i+j for i in p for j in p if i <= j] s.sort() t = bisect.bisect_left(s, m) h = bisect.bisect_left(s, m>>1, 0, t) if t < len(s): if s[t] == m: print(m); continue s = s[:t] ans = 0 # for i in range(h-1, len(s)): for i in range(h): t = bisect.bisect_left(s, m-s[i]) if s[i]+s[t-1] > ans: ans = s[i]+s[t-1] print(ans) ```
instruction
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Yes
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1
83,311
11
166,623
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem You have to play a darts game with the following rules. You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points. Create a program that finds the maximum number of points you can get given the written points and the value of M. input The input consists of multiple datasets. Each dataset is given in the following format. On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written. Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300. When both N and M are 0, the input is completed. The number of data sets does not exceed 10. output Outputs the maximum number of points you can get for each dataset on one line. Examples Input 4 50 3 14 15 9 3 21 16 11 2 0 0 Output 48 20 Input None Output None Submitted Solution: ``` import bisect # python template for atcoder1 import sys sys.setrecursionlimit(10**9) input = sys.stdin.readline def solve(): N, M = map(int, input().split()) if N*M == 0: return False P = [0]+[int(input()) for _ in range(N)] # solve k = [P[i]+P[j] for i in range(N) for j in range(i, N)] k = sorted(k) ret = 0 for tmp in k: if tmp > M: break else: r = M-tmp l = bisect.bisect_right(k, r) tmp += k[l-1] ret = max(ret, tmp) return ret ans = [] while True: ret = solve() if ret: ans.append(ret) else: break print("\n".join(map(str, ans))) ```
instruction
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Yes
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1
83,312
11
166,625
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem You have to play a darts game with the following rules. You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points. Create a program that finds the maximum number of points you can get given the written points and the value of M. input The input consists of multiple datasets. Each dataset is given in the following format. On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written. Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300. When both N and M are 0, the input is completed. The number of data sets does not exceed 10. output Outputs the maximum number of points you can get for each dataset on one line. Examples Input 4 50 3 14 15 9 3 21 16 11 2 0 0 Output 48 20 Input None Output None Submitted Solution: ``` def solve(): while True: n,m = map(int,input().split()) if not n and not m: break lst = [] lst2 = [] for i in range(n): lst.append(int(input())) for i in lst: for j in lst: lst2.append(i + j) lst2 = list(set(lst2 + lst)) lst2.sort() ans = 0 def binary(x, left, right): center = (left + right) // 2 length = right - left + 1 if length <= 0: return -100 elif length == 1: if lst2[center] > x: return center - 1 else: return center elif length == 2: if lst2[center] > x: return center - 1 else: return center elif lst2[center] == x: return x elif lst2[center] < x: return binary(x, center, right) elif lst2[center] > x: return binary(x, left, center) for i in lst2: if i <= m: ans = max(ans, i + lst2[binary(m - i, 0, len(lst2) - 1)]) else: break print(ans) solve() ```
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166,630
No
output
1
83,315
11
166,631
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. problem You have to play a darts game with the following rules. You can throw up to four arrows towards the target. You don't have to throw all four, you don't have to throw one. The target is divided into N parts. The score P1, ..., PN is written in each part. The total score S of the place where the arrow is stuck is the basis of your score. S is a predetermined score M or less. In the case of, S will be your score as it is. However, if S exceeds M, your score will be 0 points. Create a program that finds the maximum number of points you can get given the written points and the value of M. input The input consists of multiple datasets. Each dataset is given in the following format. On the first line, the integers N (1 ≤ N ≤ 1000) and M (1 ≤ M ≤ 200000000 = 2 × 108) are written in this order, separated by blanks. In (1 ≤ i ≤ N), Pi (1 ≤ Pi ≤ 100000000 = 108) is written. Of the scoring data, 20% of the points satisfy N ≤ 100 and 50% of the points satisfy N ≤ 300. When both N and M are 0, the input is completed. The number of data sets does not exceed 10. output Outputs the maximum number of points you can get for each dataset on one line. Examples Input 4 50 3 14 15 9 3 21 16 11 2 0 0 Output 48 20 Input None Output None Submitted Solution: ``` # -*- coding: utf-8 -*- """ Created on Mon May 7 22:55:31 2018 @author: maezawa """ def search(p, m, i): if i == 1: if m<p[0]: return -1 for j, k in enumerate(p): if k>m: return p[j-1] elif j==n-1: return p[n-1] else: dif = 2*10**8 for j, k in enumerate(p): if m-k<0: break a = search(p, m-k, i-1) #print(m-k, a, i-1) if m-k-a < dif: dif = m-k-a return m-dif while True: p = [] n, m = list(map(int, input().split())) if n==0 and m==0: break for i in range(n): p.append(int(input())) p.sort() dif = 2*10**8 for i in range(1,5): a = search(p, m, i) if a<0: continue #print("outer",i,a) if m-a < dif: dif = m-a print(m-dif) ```
instruction
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No
output
1
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11
166,635
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University. As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams. Notes on Test Cases Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format. When n is 0, it indicates the end of input. <!- Input n s11 s12 s13 s14 s15 ... sn1 sn2 sn3 sn4 sn5 n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100. Output Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters. Examples Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 2 100 100 100 100 100 0 0 0 0 0 1 89 90 85 93 82 0 Output 317 305 500 0 439 439 Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 Output 317 305 Input 2 100 100 100 100 100 0 0 0 0 0 Output 500 0 Input 1 89 90 85 93 82 Output 439 439 Submitted Solution: ``` while True: n = int(input()) if n == 0: break [print("{} {}".format(max(x), min(x))) for x in [[sum(map(int, input().split())) for i in range(n)]]] ```
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Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University. As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams. Notes on Test Cases Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format. When n is 0, it indicates the end of input. <!- Input n s11 s12 s13 s14 s15 ... sn1 sn2 sn3 sn4 sn5 n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100. Output Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters. Examples Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 2 100 100 100 100 100 0 0 0 0 0 1 89 90 85 93 82 0 Output 317 305 500 0 439 439 Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 Output 317 305 Input 2 100 100 100 100 100 0 0 0 0 0 Output 500 0 Input 1 89 90 85 93 82 Output 439 439 Submitted Solution: ``` while True: n = int(input()) if n == 0:break lst = [sum(map(int, input().split())) for _ in range(n)] print(max(lst), min(lst)) ```
instruction
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166,672
Yes
output
1
83,336
11
166,673
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University. As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams. Notes on Test Cases Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format. When n is 0, it indicates the end of input. <!- Input n s11 s12 s13 s14 s15 ... sn1 sn2 sn3 sn4 sn5 n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100. Output Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters. Examples Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 2 100 100 100 100 100 0 0 0 0 0 1 89 90 85 93 82 0 Output 317 305 500 0 439 439 Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 Output 317 305 Input 2 100 100 100 100 100 0 0 0 0 0 Output 500 0 Input 1 89 90 85 93 82 Output 439 439 Submitted Solution: ``` while True: n = int(input()) if n == 0: break total = [] for _ in range(n): total.append(sum(list(map(int,input().split())))) total.sort() print(total[-1],total[0]) ```
instruction
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Yes
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11
166,675
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University. As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams. Notes on Test Cases Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format. When n is 0, it indicates the end of input. <!- Input n s11 s12 s13 s14 s15 ... sn1 sn2 sn3 sn4 sn5 n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100. Output Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters. Examples Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 2 100 100 100 100 100 0 0 0 0 0 1 89 90 85 93 82 0 Output 317 305 500 0 439 439 Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 Output 317 305 Input 2 100 100 100 100 100 0 0 0 0 0 Output 500 0 Input 1 89 90 85 93 82 Output 439 439 Submitted Solution: ``` #0が来るまで回数の入力を繰り返す while True: j = int(input()) if j == 0:break #点数を合計し、一番小さいかずと、一番大きい数を出力する else: num_list = [] for _ in range(j): line = list(map(int,input().split())) num_list.append(sum(line)) print(str(max(num_list)) + " " + str(min(num_list))) ```
instruction
0
83,338
11
166,676
Yes
output
1
83,338
11
166,677
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University. As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams. Notes on Test Cases Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format. When n is 0, it indicates the end of input. <!- Input n s11 s12 s13 s14 s15 ... sn1 sn2 sn3 sn4 sn5 n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100. Output Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters. Examples Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 2 100 100 100 100 100 0 0 0 0 0 1 89 90 85 93 82 0 Output 317 305 500 0 439 439 Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 Output 317 305 Input 2 100 100 100 100 100 0 0 0 0 0 Output 500 0 Input 1 89 90 85 93 82 Output 439 439 Submitted Solution: ``` ["{} {}".format(max(x), min(x)) for x in [[sum(map(int, input().split())) for _ in range(int(input()))]]] ```
instruction
0
83,339
11
166,678
No
output
1
83,339
11
166,679
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Yui Hirasawa, who attends private Sakuragaoka Girls' High School, has to make a career hope by the day after tomorrow, but the trouble is that she hasn't decided anything yet. When I consulted with my friend Wa, I knew that my first choice was K University, so I consulted with my career guidance teacher to see if I could join K University. As a career guidance teacher who was consulted, you decided to refer to Yui's final exam results in order to predict whether Yui would be able to enter K University. However, Yui's grades are greatly affected by whether or not he wins the exam, so he decided to check the best and worst scores of the final exams in the past. The data of the past final exams only remained for each score of the five subjects, and the total score of the exam did not remain. Therefore, your job is to write a program that takes the score data of each exam as input and outputs the score of the best time and the score of the worst time in the past final exams. Notes on Test Cases Multiple datasets are given in the above input format. Create a program that outputs each data set in the above output format. When n is 0, it indicates the end of input. <!- Input n s11 s12 s13 s14 s15 ... sn1 sn2 sn3 sn4 sn5 n is the number of tests taken so far. The number of tests is 1 or more and 100 or less. The next n lines are given scores for 5 subjects in each exam. The range of points is an integer between 0 and 100. Output Output the highest and lowest points in one line. Separate the highest and lowest points with a single space and do not include any other characters. Examples Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 2 100 100 100 100 100 0 0 0 0 0 1 89 90 85 93 82 0 Output 317 305 500 0 439 439 Input 3 49 50 87 78 41 27 61 100 45 84 28 88 40 95 66 Output 317 305 Input 2 100 100 100 100 100 0 0 0 0 0 Output 500 0 Input 1 89 90 85 93 82 Output 439 439 Submitted Solution: ``` [print("{} {}".format(max(x), min(x))) for x in [[sum(map(int, input().split())) for _ in range(int(input()))]]] ```
instruction
0
83,340
11
166,680
No
output
1
83,340
11
166,681
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` n = int(input()) left = None right = None for i in range(n): st = input() op, num, ans = st.split(' ') num = int(num) if ans == 'N': if op == '>=': op = '<' elif op == '>': op = '<=' elif op == '<=': op = '>' else: op = '>=' if op == '>=': if left == None: left = num else: left = max(left, num) elif op == '>': if left == None: left = num+1 else: left = max(left, num+1) elif op == '<=': if right == None: right = num else: right = min(right, num) else: # op == '<' if right == None: right = num-1 else: right = min(right, num-1) if left == None and right == None: print(0) elif left == None: print(right) elif right == None: print(left) elif left > right: print('Impossible') else: print(left) ```
instruction
0
83,768
11
167,536
Yes
output
1
83,768
11
167,537
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` n=int(input()) l=-2*10**9 r=2*10**9 for i in range(n): s=input().split() s[1]=int(s[1]) if s[2]=='Y': if s[0]=='>=': l=max(l,s[1]) elif s[0]=='>': l=max(l,s[1]+1) elif s[0]=='<=': r=min(r,s[1]) else: r=min(r,s[1]-1) else: if s[0]=='>=': r=min(r,s[1]-1) elif s[0]=='>': r=min(r,s[1]) elif s[0]=='<=': l=max(l,s[1]+1) else: l=max(l,s[1]) if l>r: print('Impossible') else: print(l) ```
instruction
0
83,769
11
167,538
Yes
output
1
83,769
11
167,539
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` n = int(input()) questions = [] for _ in range(n): questions.append(input()) lower_bound, upper_bound = None, None for q in questions: bound = int(q.split()[1]) if (q.startswith('>') and q.endswith('Y')) or (q.startswith('<') and q.endswith('N')): if ('Y' in q and '=' not in q) or ('N' in q and '=' in q): bound += 1 lower_bound = bound if lower_bound == None else max(bound, lower_bound) else: if ('Y' in q and '=' not in q) or ('N' in q and '=' in q): bound -= 1 upper_bound = bound if upper_bound == None else min(bound, upper_bound) if lower_bound != None and upper_bound != None and lower_bound > upper_bound: print('Impossible') else: print(lower_bound if lower_bound != None else upper_bound) ```
instruction
0
83,770
11
167,540
Yes
output
1
83,770
11
167,541
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` def reverse(op): a = ['>=', '>', '<=', '<'] b = ['<', '<=', '>', '>='] indx = a.index(op) return b[indx] x = int(input()) minx = -2 * 10**9 maxx = 2 * 10**9 for i in range(x): guess = input().split(' ') if guess[2] == 'N': op = reverse(guess[0]) else: op = guess[0] comp = int(guess[1]) if op == '>=': minx = max(minx, comp) elif op == '>': minx = max(minx, comp + 1) elif op == '<=': maxx = min(maxx, comp) else: maxx = min(maxx, comp - 1) if minx > maxx: print("Impossible") else: print(minx) ```
instruction
0
83,771
11
167,542
Yes
output
1
83,771
11
167,543
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` n = int(input()) MAX = 1000000000 minimum, maximum, temp_max, temp_min = [0]*4 i = 0 while i < n: question = [x for x in input().split()] if question[2] == 'Y': if question[0] == ">=": minimum = int(question[1]) maximum = MAX elif question[0] == ">": minimum = int(question[1])+1 maximum = MAX elif question[0] == "<=": minimum = -MAX maximum = int(question[1]) else: minimum = -MAX maximum = int(question[1])-1 else: if question[0] == ">=": minimum = -MAX maximum = int(question[1]) elif question[0] == ">": minimum = -MAX maximum = int(question[1]) elif question[0] == "<=": minimum = int(question[1]) + 1 maximum = MAX else: minimum = int(question[1]) maximum = MAX if i == 0: temp_max = maximum temp_min = minimum else: temp_max = min(temp_max, maximum) temp_min = max(temp_min, minimum) if temp_max < temp_min and abs(temp_min) <= MAX and abs(temp_max) <= MAX: print("Impossible") exit() i += 1 if minimum != -MAX: print(temp_min) else: print(temp_max) ```
instruction
0
83,772
11
167,544
No
output
1
83,772
11
167,545
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` import random import math def getnum(signs, numbers, answers): number = 0 ll = min(numbers) ** 2 ul = max(numbers)**2 for i in range(0, len(signs)): if answers[i] == " Y": if signs[i] == "<": ul = numbers[i] if signs[i] == ">": ll = numbers[i] if signs[i] == ">=": if numbers[i] > ll: ll = numbers[i] + 1 if signs[i] == "<=": if numbers[i] < ul: ul = numbers[i] + 1 if answers[i] == " N": if signs[i] == "<": if numbers[i]> ll: ll = numbers[i] if signs[i] == ">": if numbers[i] < ul: ul = numbers[i] if signs[i] == ">=": if numbers[i] > ll: ul = numbers[i] + 1 if signs[i] == "<=": if numbers[i] < ul: ll = numbers[i] + 1 try: number = random.randint(ll, ul) except: number = "Impossible" print(number) number = int(input()) i = 0 ins = [] while i < number: ui = input() ins += [ui,] i += 1 signs = [] numbers = [] for i in ins: sign = i[:i.index(" ")] signs += [sign,] for k in ins: number = "" if k[2] == "-" or k[3] == "-": number += "-" for j in k: index =k.index(j) if j.isdigit(): number += j numbers += [int(number),] answers = [] for y in ins: curr = y[y.index(" ") +1 :] curr1 = curr[curr.index(" ") : ] answers += [curr1,] getnum(signs, numbers, answers) ```
instruction
0
83,773
11
167,546
No
output
1
83,773
11
167,547
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` n=int(input()) l=-(10**9)-3 r=(10**9)+3 for i in range(n): sign,x,ans=map(str,input().split()) x=int(x) if (sign==">" and ans=="Y") or (sign=="<=" and ans=="N"): if x>l: l=x-1 elif (sign=="<" and ans=="Y") or (sign==">=" and ans=="N"): if x<r: r=x+1 elif (sign==">=" and ans=="Y") or (sign=="<" and ans=="N"): if x>l: l=x elif (sign=="<=" and ans=="Y") or (sign==">" and ans=="N"): if x<r: r=x if l<=r: print(l) else: print("Impossible") ```
instruction
0
83,774
11
167,548
No
output
1
83,774
11
167,549
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show. The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions: * Is it true that y is strictly larger than number x? * Is it true that y is strictly smaller than number x? * Is it true that y is larger than or equal to number x? * Is it true that y is smaller than or equal to number x? On each question the host answers truthfully, "yes" or "no". Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible". Input The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is: * ">" (for the first type queries), * "<" (for the second type queries), * ">=" (for the third type queries), * "<=" (for the fourth type queries). All values of x are integer and meet the inequation - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no"). Consequtive elements in lines are separated by a single space. Output Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes). Examples Input 4 &gt;= 1 Y &lt; 3 N &lt;= -3 N &gt; 55 N Output 17 Input 2 &gt; 100 Y &lt; -100 Y Output Impossible Submitted Solution: ``` left_border_inclusive = float("-inf") right_border_inclusive = float("inf") inverse_signs = { ">": "<=", "<=": ">", ">=": "<", "<": ">=" } NUMBERS = int(input()) EXTRA_FINISH = False sign, value = None, None for _ in range(NUMBERS): sign, value, answer = input().split(' ') value = int(value) if answer == "N": sign = inverse_signs[sign] if sign == ">": if value >= right_border_inclusive: EXTRA_FINISH = True break left_border_inclusive = max(left_border_inclusive, value + 1) elif sign == ">=": if value > right_border_inclusive: EXTRA_FINISH = True break left_border_inclusive = max(left_border_inclusive, value) elif sign == "<": if value <= left_border_inclusive: EXTRA_FINISH = True break right_border_inclusive = max(right_border_inclusive, value-1) elif sign == "<=": if value < left_border_inclusive: EXTRA_FINISH = True break right_border_inclusive = max(right_border_inclusive, value) if not EXTRA_FINISH: print(left_border_inclusive) else: print("Impossible") ```
instruction
0
83,775
11
167,550
No
output
1
83,775
11
167,551
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy: * Each piece should contain at least l numbers. * The difference between the maximal and the minimal number on the piece should be at most s. Please help Alexandra to find the minimal number of pieces meeting the condition above. Input The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105). The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109). Output Output the minimal number of strip pieces. If there are no ways to split the strip, output -1. Examples Input 7 2 2 1 3 1 2 4 1 2 Output 3 Input 7 2 2 1 100 1 100 1 100 1 Output -1 Note For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2]. For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists. Submitted Solution: ``` n,s,l = [int(s) for s in input().split()] a = [int(s) for s in input().split()] pieces = [] m = 0 i = 1 tmpmin = a[0] tmpmax = a[0] tmppc = [a[0]] while i<n: if abs(a[i]-tmpmin)<=s and abs(a[i]-tmpmax)<=s: tmppc.append(a[i]) if a[i]<tmpmin: tmpmin=a[i] elif a[i]>tmpmax: tmpmax = a[i] else: pieces.append(tmppc) tmppc = [a[i]] tmpmin = a[i] tmpmax = a[i] i += 1 pieces.append(tmppc) fail = False for j in range(len(pieces)): if len(pieces[j])<l: if j>0: prevpc = pieces[j-1] minj = min(pieces[j]) maxj = max(pieces[j]) while len(pieces[j])<l: tmp = prevpc.pop() if abs(tmp-minj)<=s and abs(tmp-maxj)<=s: pieces[j].insert(0,tmp) else: fail = True break if len(prevpc)<l: fail = True break else: fail = True break if not fail: print(len(pieces)) else: print(-1) ```
instruction
0
83,781
11
167,562
No
output
1
83,781
11
167,563
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy: * Each piece should contain at least l numbers. * The difference between the maximal and the minimal number on the piece should be at most s. Please help Alexandra to find the minimal number of pieces meeting the condition above. Input The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105). The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109). Output Output the minimal number of strip pieces. If there are no ways to split the strip, output -1. Examples Input 7 2 2 1 3 1 2 4 1 2 Output 3 Input 7 2 2 1 100 1 100 1 100 1 Output -1 Note For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2]. For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists. Submitted Solution: ``` n,s,l = [int(s) for s in input().split()] a = [int(s) for s in input().split()] pieces = [] m = 0 i = 1 tmpmin = a[0] tmpmax = a[0] tmppc = [a[0]] while i<n: if abs(a[i]-tmpmin)<=s and abs(a[i]-tmpmax)<=s: tmppc.append(a[i]) if a[i]<tmpmin: tmpmin=a[i] elif a[i]>tmpmax: tmpmax = a[i] else: pieces.append(tmppc) tmppc = [a[i]] tmpmin = a[i] tmpmax = a[i] i += 1 pieces.append(tmppc) fail = False for j in range(len(pieces)): if len(pieces[j])<l: if j>0: prevpc = pieces[j-1] minj = min(pieces[j]) maxj = max(pieces[j]) while len(pieces[j])<l: tmp = prevpc.pop() if abs(tmp-minj)<=s and abs(tmp-maxj)<=s: pieces[j].insert(0,tmp) if tmp<minj: minj=tmp elif tmp>maxj: maxj=tmp else: fail = True break if len(prevpc)<l: fail = True break else: fail = True break if not fail: print(len(pieces)) else: print(-1) ```
instruction
0
83,782
11
167,564
No
output
1
83,782
11
167,565
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right. Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy: * Each piece should contain at least l numbers. * The difference between the maximal and the minimal number on the piece should be at most s. Please help Alexandra to find the minimal number of pieces meeting the condition above. Input The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105). The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109). Output Output the minimal number of strip pieces. If there are no ways to split the strip, output -1. Examples Input 7 2 2 1 3 1 2 4 1 2 Output 3 Input 7 2 2 1 100 1 100 1 100 1 Output -1 Note For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2]. For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists. Submitted Solution: ``` n,s,l = [int(s) for s in input().split()] a = [int(s) for s in input().split()] pieces = [] m = 0 i = 1 tmpmin = a[0] tmpmax = a[0] tmppc = [a[0]] while i<n: if abs(a[i]-tmpmin)<=s and abs(a[i]-tmpmax)<=s: tmppc.append(a[i]) if a[i]<tmpmin: tmpmin=a[i] elif a[i]>tmpmax: tmpmax = a[i] else: pieces.append(tmppc) tmppc = [a[i]] tmpmin = a[i] tmpmax = a[i] i += 1 pieces.append(tmppc) fail = False for j in range(len(pieces)): if len(pieces[j])<l: if j>0: prevpc = pieces[j-1] minj = min(pieces[j]) maxj = max(pieces[j]) while len(pieces[j])<l: tmp = prevpc.pop() if abs(tmp-minj)<=s and abs(tmp-maxj)<=s: pieces[j].append(tmp) else: fail = True break if len(prevpc)<l: fail = True break else: fail = True break if not fail: print(len(pieces)) else: print(-1) ```
instruction
0
83,783
11
167,566
No
output
1
83,783
11
167,567
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps. <image> Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two. Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps. Input The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights. The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values. Output Print the minimum number of steps in a single line. Examples Input 5 1 1 2 3 3 Output 2 Input 4 0 1 2 3 Output 4 Note In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two. In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. Submitted Solution: ``` # Description of the problem can be found at http://codeforces.com/problemset/problem/587/A n = int(input()) w = map(int, input().split()) m = 1000001 s = [0] * m for i in w: s[i] += 1 ans = 0 z = 0 for i in range(m): n_s = s[i]+z z = n_s // 2 ans += n_s % 2 while z > 0: ans += z % 2 z //= 2 print(ans) ```
instruction
0
83,803
11
167,606
Yes
output
1
83,803
11
167,607
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps. <image> Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two. Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps. Input The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights. The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values. Output Print the minimum number of steps in a single line. Examples Input 5 1 1 2 3 3 Output 2 Input 4 0 1 2 3 Output 4 Note In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two. In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. Submitted Solution: ``` n=int(input()) l=list(map(int,input().split())) cnt=[0]*(10**6+100) for i in l: cnt[i]+=1 s=0 ans=0 for i in cnt: s+=i ans+=s%2 s//=2 print(ans) ```
instruction
0
83,804
11
167,608
Yes
output
1
83,804
11
167,609
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps. <image> Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two. Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps. Input The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights. The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values. Output Print the minimum number of steps in a single line. Examples Input 5 1 1 2 3 3 Output 2 Input 4 0 1 2 3 Output 4 Note In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two. In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. Submitted Solution: ``` def power_of_two(a): return True def main(): n = int(input()) w = list(map(int, input().split())) flag = True while flag: flag = False for i in range(1, len(w)): if w[i-1] == w[i]: w[i] *= 2 del w[i-1] flag = True break print(len(w)) if __name__ == '__main__': main() ```
instruction
0
83,807
11
167,614
No
output
1
83,807
11
167,615
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps. <image> Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two. Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps. Input The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights. The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values. Output Print the minimum number of steps in a single line. Examples Input 5 1 1 2 3 3 Output 2 Input 4 0 1 2 3 Output 4 Note In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two. In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two. Submitted Solution: ``` n = int(input()) list = map(int, input().split()) N = [0]*1000001 for i in list: N[i] += 1 ans = 0 temp = 0 for j in range(1000001): number = N[j] + temp temp = number // 2 ans += number % 2 while temp > 0: number = temp temp //= 2 ans += temp%2 print(ans) ```
instruction
0
83,810
11
167,620
No
output
1
83,810
11
167,621
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` # -*- coding: utf-8 -*- import sys def input(): return sys.stdin.readline().strip() def list2d(a, b, c): return [[c] * b for i in range(a)] def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)] def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)] def ceil(x, y=1): return int(-(-x // y)) def INT(): return int(input()) def MAP(): return map(int, input().split()) def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)] def Yes(): print('Yes') def No(): print('No') def YES(): print('YES') def NO(): print('NO') sys.setrecursionlimit(10 ** 9) INF = 10 ** 18 MOD = 10 ** 9 + 7 K = INT() N = 2 ** K A = LIST(N) while len(A) >= 2: tmp = [] for i in range(len(A)//2): a, b = A[i*2], A[i*2+1] if a == b: tmp.append(a) else: tmp.append(abs(a - b)) A = tmp print(A[0]) ```
instruction
0
84,084
11
168,168
Yes
output
1
84,084
11
168,169
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` n = int(input()) a = [] for i in range(2**n): a.append(int(input())) for i in range(n): delta = 2**(i+1) half = delta // 2 for j in range(0, 2**n, delta): p = a[j]; q = a[j + half] if p == q: continue else: a[j] = abs(p - q) print(a[0]) ```
instruction
0
84,085
11
168,170
Yes
output
1
84,085
11
168,171
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` #!/usr/bin/env python3 n = int(input()) a = [ int(input()) for _ in range(2**n) ] while len(a) != 1: b = [] for i in range(0, len(a), 2): if a[i] == a[i+1]: b += [ a[i] ] else: b += [ abs(a[i] - a[i+1]) ] a = b print(*a) ```
instruction
0
84,086
11
168,172
Yes
output
1
84,086
11
168,173
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` def solve(A): if(len(A) == 1): return A[0] return solve([abs(A[i]-A[i+1]) for i in range(0,len(A),2)]) n = int(input()) A = [] for i in range(2**n): A.append(int(input())) print(solve(A)) ```
instruction
0
84,087
11
168,174
Yes
output
1
84,087
11
168,175
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` import sys N = int(sys.stdin.readline().strip()) array = [None for _ in range(2**N)] for i in range(int(2**N)): array[i] = int(sys.stdin.readline().strip()) for stage in range(N): for i in range(int(2**(N-stage-1))): array[i*2] = abs(array[i*2] - array[i*2+1]) #print(array) print(array[0]) ```
instruction
0
84,089
11
168,178
No
output
1
84,089
11
168,179
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` n=int(input()) a=[int(input())for i in range(2**n)] while len(a)!=1: s,t=a.pop(0),a.pop(0) a.append(max(s-t,t-s)) print(a[0]) ```
instruction
0
84,090
11
168,180
No
output
1
84,090
11
168,181
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kode Festival is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining. Constraints * 1 \leq N \leq 18 * 1 \leq A_i \leq 10^9 * A_i is an integer. Input The input is given from Standard Input in the following format: N A_1 A_2 : A_{2^N} Output Print the eventual hardness of the last stone remaining. Examples Input 2 1 3 10 19 Output 7 Input 3 1 3 2 4 6 8 100 104 Output 2 Submitted Solution: ``` def battle(X, Y): if X==Y: return X else: return abs(X-Y) N = int(input()) A = [] for i in range(2**N): A.append(int(input())) for i in range(2**N-1): X = A.pop(0) Y = A.pop(0) A.append(battle(X, Y)) print(A[0]) ```
instruction
0
84,091
11
168,182
No
output
1
84,091
11
168,183
Provide a correct Python 3 solution for this coding contest problem. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** *****
instruction
0
84,110
11
168,220
"Correct Solution: ``` f=0 while 1: try:n=int(input()) except:break if f:print() f=1 a=[[0 for _ in range(5)]for _ in range(8)] for i in range(4,-1,-1): b=n%10 a[not b//5][i]=1 a[b%5+3][i]=1 n//=10 for i in range(8): for j in range(5): if i==2:print('=',end='') else:print(' ' if a[i][j] else '*',end='') print() ```
output
1
84,110
11
168,221
Provide a correct Python 3 solution for this coding contest problem. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** *****
instruction
0
84,113
11
168,226
"Correct Solution: ``` # -*- coding: utf-8 -*- """ http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0128 """ import sys from sys import stdin input = stdin.readline def conv_avacus(num_txt): abacus = ['* = ****', '* =* ***', '* =** **', '* =*** *', '* =**** ', ' *= ****', ' *=* ***', ' *=** **', ' *=*** *', ' *=**** '] ans = [abacus[int(num_txt[4])], abacus[int(num_txt[3])], abacus[int(num_txt[2])], abacus[int(num_txt[1])], abacus[int(num_txt[0])]] return ans def rotate_and_print(data): y_size = len(data) x_size = len(data[0]) A = [[''] * y_size for _ in range(x_size)] for i in range(x_size): for j in range(y_size): A[i][j] = data[y_size-1-j][i] for l in A: print(''.join(l)) def main(args): first_case = True for line in sys.stdin: if not first_case: print() num_txt = line.strip().zfill(5) result = conv_avacus(num_txt) rotate_and_print(result) first_case = False if __name__ == '__main__': main(sys.argv[1:]) ```
output
1
84,113
11
168,227
Provide a correct Python 3 solution for this coding contest problem. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** *****
instruction
0
84,114
11
168,228
"Correct Solution: ``` import sys isnewLine = False pattern = {"0": "* = ****", "1": "* =* ***", "2": "* =** **", "3": "* =*** *", "4": "* =**** ", "5": " *= ****", "6": " *=* ***", "7": " *=** **", "8": " *=*** *", "9": " *=**** ", } for line in sys.stdin: if isnewLine: print() else: isnewLine = True line = line[:-1] line = "0" * (5 - len(line)) + line bar = [pattern[item] for item in line] for i1, i2, i3, i4, i5 in zip(bar[0], bar[1], bar[2], bar[3], bar[4]): print(i1 + i2 + i3 + i4 + i5) ```
output
1
84,114
11
168,229
Provide a correct Python 3 solution for this coding contest problem. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** *****
instruction
0
84,115
11
168,230
"Correct Solution: ``` f=0 while 1: try:n=int(input()) except:break if f:print() f=1 a=[['*' for _ in range(5)]for _ in range(8)] for i in range(4,-1,-1): b=n%10 a[not b//5][i]=' ' a[b%5+3][i]=' ' n//=10 a[2]='='*5 for i in a: print(''.join(i),end='') print() ```
output
1
84,115
11
168,231
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** ***** Submitted Solution: ``` f=0 while 1: try:n=int(input()) except:break if f:print() f=1 a=[[0 for _ in range(5)]for _ in range(8)] for i in range(4,-1,-1): b=n%10 if b>=5:a[0][i]=1 else: a[1][i]=1 a[b%5+3][i]=1 n//=10 for i in range(8): for j in range(5): if i==2:print('=',end='') else:print(' ' if a[i][j] else '*',end='') print() ```
instruction
0
84,116
11
168,232
Yes
output
1
84,116
11
168,233
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** ***** Submitted Solution: ``` L = [ "* = ****", "* =* ***", "* =** **", "* =*** *", "* =**** ", " *= ****", " *=* ***", " *=** **", " *=*** *", " *=**** " ] c=0 while True: try: num = int(input()) except EOFError: break if c > 0: print() c += 1 str = "{:05}".format(num) s = [int(x) for x in list(str)] for i in range(len(L[0])): t = [] for j in range(len(s)): t.append(L[s[j]][i]) print("".join(t)) ```
instruction
0
84,117
11
168,234
Yes
output
1
84,117
11
168,235
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** ***** Submitted Solution: ``` # AOJ 0128 Abacus # Python3 2018.6.19 bal4u abacus = [ [ '*', '*', '*', '*', '*', ' ', ' ', ' ', ' ', ' ' ], [ ' ', ' ', ' ', ' ', ' ', '*', '*', '*', '*', '*' ], [ '=', '=', '=', '=', '=', '=', '=', '=', '=', '=' ], [ ' ', '*', '*', '*', '*', ' ', '*', '*', '*', '*' ], [ '*', ' ', '*', '*', '*', '*', ' ', '*', '*', '*' ], [ '*', '*', ' ', '*', '*', '*', '*', ' ', '*', '*' ], [ '*', '*', '*', ' ', '*', '*', '*', '*', ' ', '*' ], [ '*', '*', '*', '*', ' ', '*', '*', '*', '*', ' ' ]] ans = [['' for c in range(5)] for r in range(8)] first = True while True: try: n = list(input()) except: break for i in range(5-len(n)): for r in range(8): ans[r][i] = abacus[r][0] for i in range(len(n)): for r in range(8): ans[r][5-len(n)+i] = abacus[r][int(n[i])] if first: first = False else: print() for r in range(8): print(*ans[r], sep='') ```
instruction
0
84,118
11
168,236
Yes
output
1
84,118
11
168,237
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. <image> At the request of a friend who started learning abacus, you decided to create a program to display the abacus beads. Create a program that takes a certain number as input and outputs a row of abacus beads. However, the number of digits of the abacus to be displayed is 5 digits, and the arrangement of beads from 0 to 9 is as follows using'*' (half-width asterisk),''(half-width blank), and'=' (half-width equal). It shall be expressed as. <image> Input Multiple test cases are given. Up to 5 digits (integer) are given on one line for each test case. The number of test cases does not exceed 1024. Output Output the abacus bead arrangement for each test case. Insert a blank line between the test cases. Example Input 2006 1111 Output **** * ===== * * **** * *** ***** ***** ***** ===== **** * ***** ***** ***** Submitted Solution: ``` def get_input(): while True: try: yield ''.join(input()) except EOFError: break table = \ ["* = ****",\ "* =* ***",\ "* =** **",\ "* =*** *",\ "* =**** ",\ " *= ****",\ " *=* ***",\ " *=** **",\ " *=*** *",\ " *=**** "] N = list(get_input()) for l in range(len(N)): if l > 0: print("") num = int(N[l]) ans = [] for i in range(5): ans.append(table[num // 10**(4-i)]) num = num % 10**(4-i) for i in range(8): for j in range(5): print(ans[j][i], end="") print("") ```
instruction
0
84,119
11
168,238
Yes
output
1
84,119
11
168,239