TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
N = int(input())
S = input()
x0 = S[0]
cnt = 1
for i in S:
    if i != x0:
        cnt += 1
    x0 = i
    print(x0)
print(cnt)

N = int(input())
S = input()
x0 = S[0]
cnt = 0
for i in S:
    if i != x0:
        cnt += 1
    x0 = i
    print(x0)
print(cnt)

#145
N = int(input())
S = input()
x0 = S[0]
cnt = 0
for i in S:
    if i != x0:
        cnt += 1
    x0 = i
print(cnt)

N = int(input())
S = input()
x0 = S[0]
cnt = 1
for i in S:
    if i != x0:
        cnt += 1
    x0 = i
print(cnt)

Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = input()

import re


def nchars(s, n):
    
    assert n > 0
    reg = re.compile("(.)\\1{%d,}" % (n - 1))  
    while True:
        m = reg.search(s)
        if not m:
            break
        yield m.group(0)
        s = s[m.end():]
        
d = nchars(s, 2)
for d_ in d:
    n -= len(d_) - 1
print(n)n = int(input())
s = input()

import re


def nchars(s, n):
    
    assert n > 0
    reg = re.compile("(.)\\1{%d,}" % (n - 1))  
    while True:
        m = reg.search(s)
        if not m:
            break
        yield m.group(0)
        s = s[m.end():]
        
d = nchars(s, 2)
for d_ in d:
    n -= len(d_) - 1
print(n)

n = int(input())
s = input()

import re


def nchars(s, n):
    
    assert n > 0
    reg = re.compile("(.)\\1{%d,}" % (n - 1))  
    while True:
        m = reg.search(s)
        if not m:
            break
        yield m.group(0)
        s = s[m.end():]
        
d = nchars(s, 2)
for d_ in d:
    n -= len(d_) - 1
print(n)

Verdicts: Runtime Error

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
N = int(input())
S = input()

total = []

def yugo(st,x):
	if len(st) >= x + 1:
		return st[x]
	if st[x] == st[x + 1]:
		return yugo(st,x+1)
	else:
		return st[x] +  yugo(st,x+1)
	

print(int(len(yugo(S,0))))


N = int(input())
S = input()

total = []
Y = 0

for i in range(N):
	if i == N-1:
		Y = Y + 1
		break
	if S[i] == S[i+1]:
		continue
	else:
		Y = Y + 1
		
print(Y) 

Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
N = str(input())
S = str(input()).split()
answer = []
 
for s in S:
  if len(answer) == 0:
    answer.append(s)
  elif answer[-1] != s:
    answer.append(s)
print(len(answer))


N = str(input())
S = list(input())
answer = []

for s in S:
  if len(answer) == 0:
    answer.append(s)
  elif answer[-1] != s:
    answer.append(s)
print(len(answer))


Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n=int(input())
t=input()
ans=t[0]
for i in range(1,n):
    if ans[-1]==i:
        pass
    else:
        ans=ans+i
print(ans)


n=int(input())
t=input()
ans=t[0]
for i in range(1,n):
    if ans[-1]==i:
        pass
    else:
        ans=ans+i
print(len(ans))

n=int(input())
t=input()
ans=t[0]
for i in range(1,n):
    if ans[-1]==t[i]:
        pass
    else:
        ans=ans+t[i]
print(len(ans))

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
# coding: utf-8
# Your code here!

N = int(input())
S = input()

print(S)

overlap = 0
for i in range(N-1) : 
    if S[i] == S[i+1] : 
        overlap += 1
print(N-overlap)

# coding: utf-8
# Your code here!

N = int(input())
S = input()

overlap = 0
for i in range(N-1) : 
    if S[i] == S[i+1] : 
        overlap += 1
print(N-overlap)

Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
N=int(input())
text=input()
x=0
for i in range(N-1):
  if text[i]!=text[i+1]:
    x+=1
print(x)


N=int(input())
text=input()
x=1
for i in range(N-1):
  if text[i]!=text[i+1]:
    x+=1
print(x)


Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = input()
res = 0
for i in range(n-1):
  if s[i] != s[i+1]:
    res += 1
print(res)

n = int(input())
s = input()
res = 1
for i in range(n-1):
  if s[i] != s[i+1]:
    res += 1
print(res)

Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
S = list(input())
ans =0
now_number =0

for i in range(n-1):
    if(S[i] != S[i+1]):
        ans +=1
print(ans)

n = int(input())
S = list(input())
ans =1
for i in range(n-1):
    if(S[i] != S[i+1]):
        ans +=1
print(ans)
    

Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
N,S=open(0)
import itertools as i
print(len(list(i.groupby(S))))

N,S=open(0)
import itertools as i
print(len(list(i.groupby(S))))


import sys
readline = sys.stdin.readline

N = int(readline())
S = readline().rstrip()

ans = 1
for i in range(1,len(S)):
  if S[i] != S[i - 1]:
    ans += 1
print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
S = input()

ans = 0

S = list(S)
for i in range(len(S)-1):
    if S[i]!=S[i+1]:
        ans += 1

print(ans)

n = int(input())
S = input()

ans = 1

S = list(S)
print(S)
for i in range(len(S)-1):
    if S[i]!=S[i+1]:
        ans =ans+1

print(ans)

n = int(input())
S = input()

ans = 1

S = list(S)
for i in range(len(S)-1):
    if S[i]!=S[i+1]:
        ans =ans+1

print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
k = list(input())

out = k[0]
for i in range(1,n):
    if k[i-1] != k[i]:
        out += k[i]
print(out)

n = int(input())
k = list(input())
out = k[0]
for i in range(1,n):
    if k[i-1] != k[i]:
        out += k[i]
print(len(out))

Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input().rstrip())
s = list(input())

for i in range(n):
    
    if s[i] == s[i+1]:
        s.pop(s[i+1])
        
print(len(s))

# coding: utf-8

n = int(input().rstrip())
l = list(input().rstrip())
count = i = 0

for i in range(1,n):
    if l[i] == l[i-1]:
        count += 1
        
print(n - count)

Verdicts: Runtime Error

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
N = int(input())
S = input()


current = ""
ans = ""
for s in S:
    if s != current:
        current = s
        ans += s

# print(len(ans))
print(ans)





N = int(input())
S = input()


current = ""
ans = ""
for s in S:
    if s != current:
        current = s
        ans += s

print(len(ans))
# print(ans)





Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
# ABC143 C
N = int(input())
S = input()
L = [s for s in S]
print(L)
M = []
cnt = 0
ans_c = 0

if N == 1:
    print(1)
    exit()

for i in range(N-1):
    if L[i]==L[i+1]:
        cnt += 1
    else:
        ans_c += cnt
        cnt = 0
    if i == N-2:
        ans_c += cnt
print(N-ans_c)

# ABC143 C
N = int(input())
S = input()
L = [s for s in S]

M = []
cnt = 0
ans_c = 0

if N == 1:
    print(1)
    exit()

for i in range(N-1):
    if L[i]==L[i+1]:
        cnt += 1
    else:
        ans_c += cnt
        cnt = 0
    if i == N-2:
        ans_c += cnt
print(N-ans_c)

Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = list(input())

i = 0
while i<len(s)-1:
  if s[i] == s[i+1]:
    del s[i]
  else:
    i += 1
print(s)


n = int(input())
s = list(input())

i = 0
while i<len(s)-1:
  if s[i] == s[i+1]:
    del s[i]
  else:
    i += 1
print("".join(s))


n = int(input())
s = list(input())

i = 0
while i<len(s)-1:
  if s[i] == s[i+1]:
    del s[i]
  else:
    i += 1
print(len(s))


Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = input()
res = ""
x = ""
for l in s:
  if i != x:
    x = l
    res += x
  else: pass
print(len(res))

n = int(input())
s = input()
res = ""
x = ""
for l in s:
  if l != x:
    x = l
    res += x
  else: pass
print(len(res))

Verdicts: Runtime Error

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
# -*- coding: utf-8 -*-
N = int(input())
S = str(input())

X = N
A = list(str(S))
print(A)
for i in range(N - 1):
    if A[i] == A[i + 1]:
        X -= 1
print(X)


# -*- coding: utf-8 -*-
n = int(input())
s = list(str(input()))

ans = 1
for i in range(n - 1):
    if s[i] != s[i + 1]:
        ans += 1
print(ans)



Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n=int(input())
a=input()
i=0
b=[a[0]]
j=0
p=0
while(i<len(a)):
    if(b[-1]!=a[i]):
        b.append(a[i])
    i=i+1
for t in b:
    print(t,end="")


n=int(input())
a=input()
i=0
b=[a[0]]
j=0
p=0
while(i<len(a)):
    if(b[-1]!=a[i]):
        b.append(a[i])
    i=i+1
print(len(b))


Verdicts: Wrong Answer

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = input()

before = aaa
ans = 0
for char in s:
  if char != before:
    before = char
    ans += 1
print(ans)


n = int(input())
s = input()
 
before = 'aaa'
ans = 0
for char in s:
  if char != before:
    before = char
    ans += 1
print(ans)

Verdicts: Runtime Error

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = input()
a = list(s)
print(a)
b = 1
while b != 0:
    b = 0
    #i = 1
    #print(i)
    for i in range(1,len(a)):
        if a[i-1] == a[i]:
            del a[i]
            print(a)
            b = b + 1
            #print(b)
            break
print(len(a))

N = int(input())
S = input()
s = S[0]
ans = 1
for i in range(1,N):
    if S[i] != s:
        ans += 1
        s = S[i]
print(ans)

Verdicts: Runtime Error

Accepted
Problem: There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.  Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.  Ultimately, how many slimes will be there?

Submission attempts (in order):
n = int(input())
s = list(input())

a = s[0]
for i in range(1,len(s)):
    if a != s[i]:
        a = s[i]
    else:
        del s[i]

print(len(s))


n = int(input())
s = list(input())

a = s[0]
ans = list(s[0])
for i in range(1,len(s)):
    if a != s[i]:
        a = s[i]
        ans.append(s[i])

print(len(ans))


Verdicts: Runtime Error

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
N = int(input())
# L = map(int, input().split())
L = [int(i) for i in input().split()]

SL = L.sort()
# print('SL', SL)
import bisect

ans = 0
for i in range(N-2):
    for j in range(i+1, N-1):
        ab = SL[i] + SL[j]
        # print(i, j)
        # print(SL[i], SL[j])
        r = bisect.bisect_left(SL, ab)
        l = j + 1
        # print('r', r)
        # print('l', l)
        # print('---')
        ans += max(0, r - l)

print(ans)

N = int(input())
L = map(int, input().split())
SL = sorted(list(L))
A = [0] * (10 ** 3 + 10)

for i in SL:
    A[i] = True
for i in range(len(A)):
    if A[i] == True:
        A[i] = A[i-1] + 1
    else:
        A[i] = A[i-1]

# print(A[4-1] - A[2])

ans = 0
for i in range(N-2):
    for j in range(i+1, N-1):
        # for k in range(j+1, N):
        #     a = SL[i]
        #     b = SL[j]
        #     c = SL[k]
        #     if a < b + c and b < c + a and c < a + b:
        #         ans += 1
        a = SL[i]
        b = SL[j]
        for k in range(j+1, N):
            c = SL[i]
            if abs(a - b) > c:
                continue
            elif c > a + b:
                break
            else:
                ans += 1
print(ans)

import bisect


N = int(input())
SL = [int(i) for i in input().split()]
SL.sort()
ans = 0
for i in range(N-2):
    for j in range(i+1, N-1):
        ab = SL[i] + SL[j]
        r = bisect.bisect_left(SL, ab)
        l = j + 1
        ans += max(0, r - l)
print(ans)


Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
N = int(input())
L = list(map(int, input().split()))
s = 0

nL = sorted(L)
print(nL)
for i in range(0, N-2):
	for j in range(i+1, N-1):
		s += sum(x<nL[i]+nL[j] for x in nL[j+1:])
		
print(s)

import bisect

N = int(input())
L = list(map(int, input().split()))
s = 0

L = sorted(L)
print(L)
for c in reversed(range(2, N)):
	for b in reversed(range(1, c)):
		a = bisect.bisect_left(L, L[c]-L[b])
		if a<b:
			s += b-a
print(s)

N = int(input())
L = list(map(int, input().split()))
s = 0

Lc = [0]*1000

for i in range(N):
	Lc[L[i]-1] += 1

for i in range(0, 998):
	if Lc[i] != 0:
		for j in range(i+1, 999):
			if Lc[j] != 0:
				for k in range(j+1, 1000):
					if Lc[k] != 0:
						if k<i+j:
							s += min(Lc[i], min(Lc[j], Lc[k]))

print(s)

import bisect

N = int(input())
L = list(map(int, input().split()))
s = 0

L = sorted(L)

for c in reversed(range(2, N)):
	for b in reversed(range(1, c)):
		a = bisect.bisect_left(L, L[c]-L[b])
		if a<b:
			s += b-a
print(s)

import bisect

N = int(input())
L = list(map(int, input().split()))
s = 0

L = sorted(L)
print(L)
for c in reversed(range(2, N)):
	for b in reversed(range(1, c)):
		a = bisect.bisect_left(L, L[c]-L[b])
		if a<b:
			s += b-a
		print("{},{},{}".format(a,b,c))

print(s)

import bisect

N = int(input())
L = list(map(int, input().split()))
s = 0

L = sorted(L)

for c in reversed(range(2, N)):
	for b in reversed(range(1, c)):
		a = bisect.bisect_right(L, L[c]-L[b])
		if a<b:
			s += b-a
print(s)

Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
n = int(input())
a = sorted(map(int, input().split()))
r = 0
for i in range(2, n):
    j, k = 0, i - 1
    while j <= k:
        while j <= k and a[j] + a[k] > a[i]: k -= 1
        r += i - k - 1
        j += 1
print(r)


from bisect import bisect_right
 
n = int(input())
a = sorted(map(int, input().split()))
print(sum(j - bisect_right(a, a[i] - a[j], 0, j)
          for i in range(2, n) for j in range(1, i)))

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect
N = int(input())
L = [int(i) for i in input().split()]

SL = sorted(L)
ans = 0
for i in range(N - 2):
    for j in range(i + 1, N - 1):
        right = bisect.bisect_left(SL, SL[i] + SL[j])
        left = j + 1
        ans += max(right - left)

print(ans)


import bisect
N = int(input())
L = [int(i) for i in input().split()]

SL = sorted(L)
ans = 0
for i in range(N - 2):
    for j in range(i + 1, N - 1):
        right = bisect.bisect_left(SL, SL[i] + SL[j])
        left = j + 1
        ans += right - left

print(ans)


Verdicts: Runtime Error

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
)import bisect
n = int(input())
l = sorted(list(map(int,input().split())))
ans = 0
 
for i in range(n-2):
    for j in range(i+1, n-1):
        ans += bisect.bisect_left(l,l[i] + l[j]) - j - 1
print(ans)

l = list(map(int, input().split()))
l = sorted(l)
ans = 0
for i in range(n):
  a = l[i]
  for j in range(i + 1, n):
    b = l[j]
    bel1 = j #OK
    up1 = n #OUT
    while up1- bel1 > 1:
      mid = (bel1 + up1)//2
      if a + b > l[mid]:
        bel1 = mid
      else:
        up1 = mid
    ans += (bel1-j)
print(ans)

import bisect
n = int(input())
l = sorted(list(map(int, input().split())))
ans = 0
for i in range(n-2):
  a = l[i]
  for j in range(i + 1, n-1):
    ans += bisect.bisect_left(l, l[j]+a)-j-1
print(ans)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
n = int(input())
l = list(map(int, input().split()))
l.sort()
c = 0
for i in range(n - 2):
    x = l[i]
    for j in range(i + 1, n - 1):
        y = l[j]
        for k in range(j + 1, n):
            if x + y <= l[k]:
                break    
        c += k - j - 1
print(c)

import bisect

n = int(input())
l = list(map(int, input().split()))
l.sort()
c = 0
for i in range(n - 2):
    for j in range(i + 1, n - 1):   
        c += bisect.bisect_left(l, l[i] + l[j]) - j - 1
print(c)

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
n=int(input())
L=list(map(int,input().split()))
L.sort(reverse=True)
ans=0
for i in range(n):
  for j in range(i+1,n):
    for k in range(j+1,n):
      if L[i] > L[j] + L[k]:
        ans+=k-j-1
        break
      ans+=1
print(ans)

import bisect
n=int(input())
L=sorted(list(map(int,input().split())))
ans=0
for i in range(n-2):
  for j in range(i+1,n-1):
    k=bisect.bisect_left(L,L[i]+L[j],j+1)-1
    ans+=k-j
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
N = int(input())
Ls = list(map(int, input().split()))
Ls.sort()
print(Ls)
count = 0
for i in range(N-2):
    a = Ls[i]
    for k in range(i+1,N-1):
        ab = a + Ls[k]
        count += len(list(filter(lambda c: c<ab, Ls[k+1:N])))
print(count)

import bisect
n = int(input())
lis = sorted(list(map(int,input().split())))
count = 0
for i in range(n-2):
    for j in range(i+1,n-1):
        index = bisect.bisect_left(lis,lis[i]+lis[j])
        count += index-1-j
print(count)

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)

count = 0
for n in range(N-2):
    for m in range(n+1,N-1):
        print(n,m)
        print(bisect.bisect_left(L,L[m]+L[n]))
        count += bisect.bisect_left(L,L[m]+L[n]) -m-1

print(count)



import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)

count = 0
for n in range(N-2):
    for m in range(n+1,N-1):
        count += N - bisect.bisect_left(L,L[m]+L[n])

print(count)


import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)

print(L)

count = 0
for n in range(N-2):
    for m in range(n+1,N-1):
        count += N - bisect.bisect_left(L,L[m]+L[n])

print(count)


import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)

count = 0
for n in range(N-2):
    for m in range(n+1,N-1):
        count += bisect.bisect_left(L,L[m]+L[n]) - (m + 1)

print(count)


Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
N = int(input())
A = list(map(int, input().split()))
A.sort()
from bisect import bisect_left, bisect_right
ans = 0
for i in range(N - 1):
    for j in range(i + 1, N):
        if bisect_left(A, A[i] + A[j]) != bisect_right(A, A[i] + A[j]):
            ans += bisect_left(A, A[i] + A[j]) - bisect_right(A, A[j]) - 1
        else:
            ans += bisect_left(A, A[i] + A[j]) - bisect_right(A, A[j])
print(ans)

N = int(input())
A = list(map(int, input().split()))
from bisect import bisect_left
ans = 0
for i in range(N - 1):
    for j in range(i + 1, N):
        ans += bisect_left(A, A[i] + A[j]) - bisect_left(A, A[j]) - 1
print(ans)

N = int(input())
A = sorted(list(map(int, input().split())))
ans = 0
for i in range(N):
    k = i
    for j in range(i + 1, N):
        while k < N and A[k] < A[j] + A[i]:
            k += 1
        ans += (k - 1) - j
print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect


def main():
    n = int(input())
    ls = sorted(list(map(int, input().split())))

    ans = 0
    for i in range(n-1):
        for j in range(i+1, n):
            a = ls[i]
            b = ls[j]
            k = bisect.bisect_left(ls, a+b)
            if k > j:
                ans += 1
    print(ans)


if __name__ == "__main__":
    main()


import bisect


def main():
    n = int(input())
    ls = sorted(list(map(int, input().split())))

    ans = 0
    for i in range(n-1):
        for j in range(i+1, n):
            a = ls[i]
            b = ls[j]
            k = bisect.bisect_left(ls, a+b)
            if k > j:
                ans += k - j - 1
    print(ans)


if __name__ == "__main__":
    main()


Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect

n = int(input())
l = sorted([int(i) for i in input().split()])
ans = 0
for d in range(n):
    for e in range(d+1, n):
        a, b = l[d], l[e]
        idx = bisect.bisect_left(l, a + b)
        ans += max(idx - e - 1,)
print(ans)


import bisect

n = int(input())
l = sorted([int(i) for i in input().split()])
ans = 0
for d in range(n):
    for e in range(d+1, n):
        a, b = l[d], l[e]
        idx = bisect.bisect_left(l, a + b)
        ans += max(idx - e - 1, 0)
print(ans)


Verdicts: Runtime Error

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
from itertools import combinations
from collections import defaultdict

N = int(input())
L = list(map(int, input().split()))
L_ = sorted(L)
d = defaultdict(int)

for l in L_:
    d[l] += 1
n = 0
for i in range(2001):
    n += d[i]
    d[i] += n

ans = 0

for a, b in combinations(L_, 2):
    n =  d[a+b-1] - d[abs(a-b)]
    if abs(a-b) < a:
        ans -= 1
    if abs(a-b) < b:
        ans -= 1
    ans += n
    
print(ans//3)


from itertools import combinations
from collections import defaultdict

N = int(input())
L = list(map(int, input().split()))
L_ = sorted(L)
d = defaultdict(int)

for l in L_:
    d[l] += 1

n = 0
for i in range(2001):
    n += d[i]
    d[i] += n - d[i]

ans = 0

for a, b in combinations(L_, 2):
    #print(a, b, d[a+b-1], d[abs(a-b)], d[a+b-1] - d[abs(a-b)])
    n =  d[a+b-1] - d[abs(a-b)]
    if abs(a-b) < a:
        ans -= 1
    if abs(a-b) < b:
        ans -= 1
    ans += n
    
print(ans//3)


Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect
N = int(input())
L = [int(x) for x in input().split()]
L = sorted(L)
ans = 0
for one in range(N - 2):
    for two in range(one+1, N-1):
        ans += bisect.bisect_left(L, (L[one] + L[two]))-two-1
        print(bisect.bisect_left(L, (L[one] + L[two]))-two-1)
print(ans)


import bisect
N = int(input())
L = [int(x) for x in input().split()]
L = sorted(L)
ans = 0
for one in range(N - 2):
    for two in range(one+1, N-1):
        ans += bisect.bisect_left(L, (L[one] + L[two]))-two-1
        
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect
N=int(input())
print(N)
L=sorted(list(map(int,input().split())))

ans=0
for i in range(N-1):
  for j in range(i+1,N):
    
    ans+=bisect.bisect_left(L,L[i]+L[j],j+1)-j-1
print(ans)

import bisect
N=int(input())
print(N)
L=sorted(list(map(int,input().split())))

ans=0
for i in range(N-2):
  for j in range(i+1,N-1):
    
    ans+=bisect.bisect_left(L,L[i]+L[j],j+1)-j-1
print(ans)

import bisect
N=int(input())
L=sorted(list(map(int,input().split())))

ans=0
for i in range(N-2):
  for j in range(i+1,N-1):
    # ans+=bisect.bisect_left(L,L[i]+L[j])-j-1 # Code 1 
    ans+=bisect.bisect_left(L,L[i]+L[j],j)-j-1 # Code 2 
print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
N=int(input())
num=list(map(int, input().split()))
c=0
for i in range(0,N-3,1): 
    for j in range(i+1,N-2,1):
        for k in range(i+2,N-1,1):
            t=int((num[i]+num[j]+num[k])/2)
            if num[i]<t and num[j]<t and num[k]<t:
                c+=1
print (c)
       

import bisect
N=int(input())
num=list(map(int, input().split()))
c=0
num.sort()
for i in range(N-1,1,-1): 
    for j in range(i-1,0,-1):
        t=num[i]-num[j]
        k=bisect.bisect_right(num,t,0,j)
        c+=(j-k)
print (c)
       

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)
ct = 0

for i in range(N-1):
    a = L[i]
    for j in range(i+1, N):
        b = L[j]
        up_bound = bisect.bisect_right(L, a+b)
        ct += max(0, up_bound - (j+1))

print(ct)


import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)
ct = 0

for i in range(N-1):
    a = L[i]
    for j in range(i+1, N):
        b = L[j]
        up_bound = bisect.bisect_right(L, a+b)
        ct += up_bound - (j+1)

print(ct)


import bisect

N = int(input())
L = list(map(int, input().split()))
L = sorted(L)
ct = 0

for i in range(N-1):
    a = L[i]
    for j in range(i+1, N):
        b = L[j]
        up_bound = bisect.bisect_left(L, a+b)
        ct += max(0, up_bound - (j+1))

print(ct)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import bisect
n = int(input())
l = List(map(int, input().split()))
count = 0
l = sorted(l)

for i in range(n-2):
  for j in range(i+1, n-1):
    k = bisect.bisect_left(l, l[i]+l[j])
    count += k -j - 1
print(count)

import bisect
n = int(input())
l = list(map(int, input().split()))
count = 0
l = sorted(l)

for i in range(n-2):
  for j in range(i+1, n-1):
    k = bisect.bisect_left(l, l[i]+l[j])
    count += k -j - 1
print(count)

Verdicts: Runtime Error

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
import math
import collections
import fractions
import itertools
import functools
import operator
import bisect

def solve():
    n = int(input())
    l = list(map(int, input().split()))
    l.sort()
    res = 0
    for i in range(n):
        for j in range(i+1, n):
            k = bisect.bisect(l, l[i]+l[j])
            res += max(k-(j+1), 0)
    print(res)
    return 0

if __name__ == "__main__":
    solve()

import math
import collections
import fractions
import itertools
import functools
import operator
import bisect

def solve():
    n = int(input())
    l = list(map(int, input().split()))
    l.sort()
    res = 0
    for i in range(n):
        k = i
        for j in range(i+1, n):
            while(k < n and l[k] < l[i]+l[j]):
                k += 1
            res += k - (j+1)
    print(res)
    return 0

if __name__ == "__main__":
    solve()

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
# ABC143D - Triangles
def binary_search(cur: int, upper_bound: int, lower_bound: int) -> int:
    ok, ng, ret = upper_bound, lower_bound, 0
    while abs(ok - ng) > 1:
        if L[ok] + L[ng] > cur:
            ret += ok - ng
            ok -= 1
        else:
            ng += 1
    return ret


def main():
    # fix the longest edge -> binary search
    global L
    N, *L = map(int, open(0).read().split())
    L.sort()
    ans = sum(binary_search(L[i], i - 1, 0) for i in range(N - 1, 1, -1))
    print(ans)


if __name__ == "__main__":
    main()

# ABC143D - Triangles
def main():
    # fix one of edges -> binary search
    N, *L = map(int, open(0).read().split())
    L.sort()
    ans = 0
    for i in range(N - 1, 1, -1):
        x, ok, ng = L[i], 0, i - 1
        while ng < ok:
            if L[ng] + L[ok] > x:
                ans += ok - ng
                ok -= 1
            else:
                ng += 1
    print(ans)


if __name__ == "__main__":
    main()

# ABC143D - Triangles
def binary_search(cur: int, upper_bound: int, lower_bound: int) -> int:
    ok, ng, ret = upper_bound, lower_bound, 0
    while abs(ok - ng) >= 1:
        if L[ok] + L[ng] > cur:
            ret += ok - ng
            ok -= 1
        else:
            ng += 1
    return ret


def main():
    # fix the longest edge -> binary search
    global L
    N, *L = map(int, open(0).read().split())
    L.sort()
    ans = sum(binary_search(L[i], i - 1, 0) for i in range(N - 1, 1, -1))
    print(ans)


if __name__ == "__main__":
    main()

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.  He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:    * a < b + c   * b < c + a   * c < a + b  How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Submission attempts (in order):
N=int(input())
L=list(map(int, input().split()))
L.sort()
cnt=0
first=True
left=2

for i in range(0,N-2): 
  for j in range(i+1,N-1):
    for k in range(left,N):
      if L[k]<L[i]+L[j]:
        if L[i]<L[j]+L[k]:
          if L[j]<L[i]+L[k]:
            
            cnt+=1
            if first==True:
              left=k
              first=False
  left=j+1
      
print(cnt)

import bisect

N=int(input())
L=list(map(int, input().split()))
L.sort()

ans=0
for i in range(N-2):
  for j in range(i+1,N-1):
    
    large = bisect.bisect_left(L, L[j]+L[i])    
    #print(i,j,L[i],L[j],"  ",L[j]-L[i],L[j]+L[i],small,large)
    ans += max(large-j-1,0)    

print(ans)

Verdicts: Wrong Answer

Accepted