TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
xs,ys,xt,yt = map(int, input().split())
n = int(input())
c = [(xs,ys, 0),(xt,yt,0)]
for i in range(n):
    x,y,r = map(int, input().split())
    c.append((x,y,r))

def circle_circle(ax,ay,ar,bx,by,br):
    d =((ax-bx)**2 + (ay-by)**2)**0.5
    return max(0, d-(ar+br))

def dijkstra(V, edges, root):
    import heapq

    visited = [False] * V
    nodes = [[] for i in range(V)]
    cost = [[] for i in range(V)]

    for e in edges:
        s, t, d = e
        nodes[s].append(t)
        cost[s].append(d)

    dist = [float('inf')] * V
    dist[root] = 0

    h = [(0, root)]
    while h:
        c, v = heapq.heappop(h)
        if visited[v]:
            continue
        visited[v] = True
        for i in range(n+1):
            t = nodes[v][i]
            nc = c + cost[v][i]
            if dist[t] > nc:
                dist[t] = nc
                heapq.heappush(h, (nc, t))
    return dist
edges = []
for i in range(n+1):
    for j in range(i+1, n+2):
        d = circle_circle(c[i][0],c[i][1],c[i][2],c[j][0],c[j][1],c[j][2])
        edges.append((i,j,d))
        edges.append((j,i,d))
dij = dijkstra(n+2, edges, 0)
print(dij[1])


n,x = map(int, input().split())
a = list(map(int, input().split()))
b = a[::-1]

def solve(a):
    total = 0
    if a[0] > x:
        total += a[0]-x
        a[0] = x

    for i in range(1, n):
        if a[i-1] + a[i] > x:
            d = (a[i-1] + a[i]) - x
            total += d
            a[i] = max(0, a[i] - d)
    return total

print(min(solve(a), solve(b)))


Verdicts: Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
from math import sqrt
inf = 10**10
x_s, y_s, x_t, y_t = map(int, input().split())
N = int(input())
xyr = []
for i in range(N):
	x_i, y_i, r_i = map(int, input().split())
	xyr.append((x_i, y_i, r_i))
p = (x_s, y_s)
goal = (x_t, y_t)
ans = 0
while p != goal and xyr:
	D = inf
	for i in range(N):
		x, y, r = xyr[i]
		d = sqrt((x-p[0])**2+(y-p[1])**2) - 2*r
		if d < D:
			j = i
			D = d
	d = sqrt((goal[0]-p[0])**2+(goal[1]-p[1])**2)
	if d < D:
		ans += d
		p = goal
	else:
		x, y, r = xyr.pop(j)
		p = (x, y)
		N -= 1
		ans += D

if p != goal:
	d = sqrt((goal[0]-p[0])**2+(goal[1]-p[1])**2)
	ans += d

print(max(0, ans))
	

N, x = map(int, input().split())
a = [int(i) for i in input().split()]
ans = 0
for i in range(N-1):
	if a[i] + a[i+1] > x:
		d = a[i] + a[i+1] - x
		if d <= a[i+1]:
			a[i+1] -= d
			ans += d
		else:
			a[i] -= d - a[i+1]
			a[i+1] = 0
			ans += d
print(ans)

Verdicts: Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
def getinputdata():

    
    array_result = []
    
    data = input()
    
    array_result.append(data.split(" "))

    flg = 1

    try:

        while flg:

            data = input()

            if(data != ""):
                
                array_result.append(data.split(" "))

                flg = 1

            else:

                flg = 0
    finally:

        return array_result

arr_data = getinputdata()

n = int(arr_data[0][0])
x = int(arr_data[0][1])

arr=[int(arr_data[1][x]) for x in range(0,n)]
cnt=0
for i in range(n-1):
    
    if arr[i]+arr[i+1]>x:
        
        if arr[i]-x>=0:
            
            cnt+=arr[i-x]
            arr[i]=0
            
        if arr[i]+arr[i+1]-x>=0:
            
            cnt+=arr[i+1]-x
            arr[i+1]=arr[i+1]-x
                   
print(cnt)        
 

def getinputdata():

    
    array_result = []
    
    data = input()
    
    array_result.append(data.split(" "))

    flg = 1

    try:

        while flg:

            data = input()

            if(data != ""):
                
                array_result.append(data.split(" "))

            else:

                flg = 0
    finally:

        return array_result

arr_data = getinputdata()

n = int(arr_data[0][0])
x = int(arr_data[0][1])

arr=[int(arr_data[1][x]) for x in range(0,n)]

cnt=0

for i in range(n-1):
    
    zan=arr[i]+arr[i+1]-x
     
    if 0<arr[i+1]<=zan:

        cnt+=arr[i+1]
        zan-=arr[i+1]
        arr[i+1]=0

    elif arr[i+1]>zan>0:

        cnt+=zan
        arr[i+1]-=zan
        zan=0

    if 0<arr[i]<=zan:

        cnt+=arr[i]
        zan-=arr[i]
        arr[i]=0

    elif arr[i]>zan>0:

        cnt+=zan
        arr[i]-=zan
        zan=0

    
print(cnt)    

Verdicts: Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
def dijkstra(s):
    
    
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0
    
    while True:
        v = -1
        
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True
               
        for j in range(n):
               d[j] = min(d[j],d[v]+cost[v][j])
    return d

Xs,Ys,Xt,Yt = map(int,input().split())
n = int(input())
p = [[Xs,Ys,0],[Xt,Yt,0]]
for i in range(2,n+2):
    p.append(list(map(int,input().split())))
cost = [[0]*(n+2)for i in range(n+2)]

for i in range(n+2):
    for j in range(i,n+2):
        cost[i][j] = cost[j][i] =  max(((p[i][0]-p[j][0])**2+(p[i][1]-p[j][1])**2)**0.5 - p[i][2] - p[j][2],0)

n += 2

res = dijkstra(0)
print(res[1])


n,x = map(int,input().split())
a = list(map(int,input().split()))

res = 0
for i in range(1,n):
    if a[i]+a[i-1]>= x:
        res += (a[i]+a[i-1]-x)
        if a[i-1]>= x:
            a[i]=0
        else:
            a[i] = x-a[i-1]
    
print(res)

Verdicts: Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
N, x = map(int, input().split())
l = list(map(int, input().split()))

before = sum(l)
l[0] = min(x, l[0])
for i in range(1, N):
    s = l[i-1]+l[i]
    if s > x:
        l[i] = x-l[i-1]

print(l)
print(before-sum(l))


from collections import defaultdict
import heapq
from math import sqrt
# -*- coding: utf-8 -*-
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline


def calc(b1, b2):
    x1, y1, r1 = b1
    x2, y2, r2 = b2
    d = sqrt((x1-x2)**2+(y1-y2)**2)
    return max(d-(r1+r2), 0)


sx, sy, gx, gy = map(int, input().split())
N = int(input())
barriers = [[sx, sy, 0]] + [list(map(int, input().split()))
                            for _ in range(N)]+[[gx, gy, 0]]

mat = [[0]*(N+2) for _ in range(N+2)]

for row in range(N+2):
    for col in range(row+1, N+2):
        d = calc(barriers[row], barriers[col])
        mat[row][col] = d
        mat[col][row] = d


dist = [float('inf')]*(N+2)
dist[0] = 0


prev = defaultdict(lambda: None)

Q = []
# Q->[dist from　start,node])
heapq.heappush(Q, (0, 0))

while Q:
    dist_to_node, node = heapq.heappop(Q)
    if dist[node] < dist_to_node:
        continue
    else:
        for v in range(N+2):
            weight = mat[node][v]
            alt = dist_to_node+weight
            if dist[v] > alt:
                dist[v] = alt
                prev[v] = node
                heapq.heappush(Q, (alt, v))
print(dist[-1])


N, x = map(int, input().split())
l = list(map(int, input().split()))

before = sum(l)
l[0] = min(x, l[0])
for i in range(1, N):
    s = l[i-1]+l[i]
    if s > x:
        l[i] = x-l[i-1]

print(before-sum(l))


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

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

ans=0
if s[0]>x:
    print(s[0])
    ans=s[0]-x
    s[0]=x
for i in range(len(s)-1):
    if s[i]+s[i+1]>x:
        ans+=s[i]+s[i+1]-x
        print(ans)
        s[i+1]=x-s[i]
print(ans)       
        

N,x = map(int,input().split())
a = list(map(int,input().split()))
count = 0
for i in range(N-1):
    if a[i]+a[i+1] >= x:
        tmp = (a[i]+a[i+1]-x)
        a[i+1] -= tmp
        a[i+1] = max(0,a[i+1])
        count += tmp
print(count)

Verdicts: Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
n, x = map(int, input().split())
l = list(map(int, input().split()))
total = 0

for i in range(n - 1):
    if l[i] + l[i + 1] > x:
        max_index = None
        if l[i] > l[i+1]:
            max_index = i
            o = i+1
        else:
            max_index = i+1
            o = i
        l[max_index] = x-l[o]
        total += x-l[o]



n, x = map(int, input().split())
l = list(map(int, input().split()))
total = 0

for i in range(n - 1):
    diff = l[i] + l[i+1] - x
    if diff > x:
        total += diff
        if diff > l[i+1]:
            l[i+1]=0
        else:
            l[i+1] -= diff

print(total)


n, x = map(int, input().split())
l = list(map(int, input().split()))
total = 0

for i in range(n - 1):
    diff = l[i] + l[i+1] - x
    if diff > 0:
        total += diff
        if diff > l[i+1]:
            l[i+1]=0
        else:
            l[i+1] -= diff

print(total)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
n,k = map(int, input().split())
l = list(map(int, input().split()))
tot = 0
for i in range(1,n):
  if a[i-1] + a[i] > k:
    c = min(a[i],k-(a[i]+a[i-1]))
    tot += c
    a[i] -= c
    if a[i-1]+a[i] > k:
      d = k-a[i-1]
      tot += d
      a[i-1] -= d
print(tot)

n,k = map(int, input().split())
a = list(map(int, input().split()))
tot = 0
for i in range(1,n):
  if a[i-1] + a[i] > k:
    c = min(a[i],k-(a[i]+a[i-1]))
    tot += c
    a[i] -= c
    if a[i-1]+a[i] > k:
      d = k-a[i-1]
      tot += d
      a[i-1] -= d
print(tot)

n,k = map(int, input().split())
a = list(map(int, input().split()))
tot = 0
for i in range(1,n):
  if a[i-1] + a[i] > k:
    c = min(a[i],a[i]+a[i-1]-k)
    tot += c
    a[i] -= c
    if a[i-1]+a[i] > k:
      d = a[i-1]-k
      tot += d
      a[i-1] -= d
print(tot)

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
# ARC 064
def getInt(): return int(input())
def getIntList(): return [int(x) for x in input().split()]
def db(x): 
    global debug
    if debug: print(x)
debug = False

N,x = getIntList()
a = getIntList()
eatCnt = 0
half = (x+1) // 2 
db((N,x,a,half))
for i in range(N-1):
    if a[i]+a[i+1]>x: 
        #eat = a[i]+a[i+1]-(min(a[i],half)+min(a[i+1],half))
        eat = a[i] - min(a[i],max(a[i+1]-x,0),half)
        eatCnt += eat
        a[i] = min(a[i],max(a[i+1]-x,0),half)
eat = a[N-1] - min(a[N-1],half)
eatCnt += eat
a[N-1] = min(a[N-1],half)
db((eatCnt,a))
i=0
while i<N:
    while i<N-1 and a[i]+a[i+1]<=x: i+=1
    cnt=1
    while i<N-1 and a[i]+a[i+1]>x: 
        i+=1; cnt+=1
    eatCnt += cnt//2
    db((cnt,eatCnt))
    i+=1
print(eatCnt)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Feb 11 09:54:00 2019

@author: shinjisu
"""


# ARC 064 WA
def getInt(): return int(input())


def getIntList(): return [int(x) for x in input().split()]


def dmp(x):
    global debug
    if debug:
        print(x)


def probC_try1():
    N, x = getIntList()
    a = getIntList()
    dmp((N, x, a))
    eatCnt = 0
    half = (x+1) // 2
    dmp((N,x,a,half))
    for i in range(N-1):
        if a[i]+a[i+1] > x:
            #eat = a[i]+a[i+1]-(min(a[i],half)+min(a[i+1],half))
            eat = a[i] - min(a[i],max(a[i+1]-half,half),half)
            eatCnt += eat
            a[i] = min(a[i],max(a[i+1]-half,half),half)
    eat = a[N-1] - min(a[N-1], half)
    eatCnt += eat
    a[N-1] = min(a[N-1],half)
    dmp((eatCnt,a))
    i=0
    while i<N:
        while i<N-1 and a[i]+a[i+1]<=x: i+=1
        cnt=1
        while i<N-1 and a[i]+a[i+1]>x: 
            i+=1; cnt+=1
        eatCnt += cnt//2
        dmp((cnt,eatCnt))
        i+=1
    return eatCnt


def probC():
    N, X = getIntList()
    a = getIntList()
    dmp((N, X, a))
    eatCnt = 0
    for i in range(N-1):
        dmp((eatCnt,a))
        if a[i] > X:
            eat = a[i] - X
            dmp(('eat',eat))
            eatCnt += eat
            a[i] = X
        if a[i]+a[i+1] > X:
            eat = a[i]+a[i+1] - X
            dmp(('eat',eat))
            eatCnt += eat
            if a[i+1] < eat:
                # a[i] -= X-a[i+1]
                a[i+1] = 0
            else:
                a[i+1] -= eat
    dmp((eatCnt,a))
    return eatCnt


debug = False  # True False
print(probC())


Verdicts: Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
n,x=map(int,input().split())
a=list(map(int,input().split()))
ret=0
if a[0]>x:
  ret+=a[0]-x
  a[0]-=x
for i in range(1:n):
  if a[i]+a[i-1]>x:
    ret+=a[i]+a[i-1]-x
    a[i]=a[i]+a[i-1]-x
print(ret)

n,x=map(int,input().split())
a=list(map(int,input().split()))
ret=0
if a[0]>x:
  ret+=a[0]-x
  a[0]-=a[0]-x
for i in range(n-1):
  if a[i+1]+a[i]>x:
    ret+=a[i+1]+a[i]-x
    a[i+1]-=a[i+1]+a[i]-x
print(ret)

Verdicts: Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
# -*- coding: utf-8 -*-
line = input().split(" ")
n = int(line[0])
x = int(line[1])
a = [int(n) for n in input().split(" ")]

ret = max(0,a[0]-x)
a[0] = min(a[0],x)
for i in range(1,n):
    if a[i-1]+a[i]>x:
        r = (a[i-1]+a[i])-x
        if a[i]>=r:
            a[i] -= r
        else:
            a[i-1] -= (r-a[i])
            a[i] = 0
        ret += r

# print(ret)
print(a)


n,x = map(int, input().split())
al = list(map(int, input().split()))

res = 0
for i in range(n-1):
    a1 = al[i]
    a2 = al[i+1]
    if a1+a2>x:
        d = a1+a2-x
        res += d
        if a2>=d:
            al[i+1] -= d
        else:
            al[i+1] = 0

print(res)

Verdicts: Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
import sys

n, x, *lst = map(int, sys.stdin.read().split())
re1 = 0
lst.append(0)
for i in range(n, 0, -1):
  tmpsum = lst[i] + lst[i - 1]
  if tmpsum > x:
    res += tmpsum - x
    lst[i - 1] -= tmpsum - x

print(res)


import sys

n, x, *lst = map(int, sys.stdin.read().split())
res = 0
lst.append(0)
for i in range(n, 0, -1):
  tmpsum = lst[i] + lst[i - 1]
  if tmpsum > x:
    res += tmpsum - x
    lst[i - 1] -= tmpsum - x

print(res)


Verdicts: Runtime Error

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
from sys import stdin, stdout
import numpy as np
def readLine_int_list():return list(map(int, stdin.readline().split()))



def main():
    n, x = readLine_int_list()

    _a = readLine_int_list()
    a = _a.copy()
    
    for i in range(n-1):
        if a[i] + a[i+1] > x:
            
            if a[i] > x:
                a[i] = x
            
            if x - a[i] >= 0:
                if a[i] == 0:
                    a[i+1] = x
                elif a[i] > 0:
                    a[i+1] = 0
    print(np.sum(np.array(_a) - np.array(a)))
    
if __name__ == '__main__':\
    main()


from sys import stdin, stdout
import numpy as np
def readLine_int_list():return list(map(int, stdin.readline().split()))



def main():
    n, x = readLine_int_list()

    _a = readLine_int_list()
    a = _a.copy()
    
    for i in range(n-1):
        if a[i] + a[i+1] > x:
            
            if a[i] > x:
                a[i] = x
            
            if x - a[i] >= 0:
                if a[i] == 0:
                    a[i+1] = x
                elif a[i] > 0:
                    a[i+1] = x-a[i]
    print(np.sum(np.array(_a) - np.array(a)))
    
if __name__ == '__main__':
    main()


Verdicts: Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
n,x = (int(i) for i in input().split())
a = list(int(i) for i in input().split())
copy = []
for num in a:
    copy.append(num)
    
for i in range(n-1):
    if i==0:
        if a[i]>x:
            a[i]=x
    else:
        if a[i]+a[i+1]>x:
            a[i+1] -= x-(a[i]+a[i+1])

print(sum(copy)-sum(a))


n,x = (int(i) for i in input().split())
a = list(int(i) for i in input().split())
copy = []
for num in a:
    copy.append(num)
    
for i in range(n-1):
    if i==0:
        if a[i]>x:
            a[i]=x
    if a[i]+a[i+1]>x:
        a[i+1] = x-a[i]

print(sum(copy)-sum(a))


Verdicts: Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
N, x = map(int, input().split())
A = list(map(int, input().split()))

ans = 0

for i in range(N - 1):
    if A[i] + A[i + 1] > x:
        ans += A[i] + A[i + 1] - x
        if A[i] >= A[i + 1]:
            d = A[i] + A[i + 1] - x
            if d < 0:
                A[i] = 0
                A[i + 1] -= x - A[i]
            else:
                A[i] -= d

        else:
            d = A[i] + A[i + 1] - x
            if d < 0:
                A[i + 1] = 0
                A[i] -= x - A[i + 1]
            else:
                A[i + 1] -= d

print(ans)


N, x = map(int, input().split())
A = list(map(int, input().split()))

ans = 0

for i in range(N - 1):
    if A[i] + A[i + 1] > x:
        d = A[i] + A[i + 1] - x
        ans += d
        if A[i + 1] - d < 0:
            A[i] -= d - A[i + 1]
            A[i + 1] = 0
        else:
            A[i + 1] -= d

print(ans)


Verdicts: Wrong Answer

Accepted
Problem: There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.  Snuke can perform the following operation any number of times:    * Choose a box containing at least one candy, and eat one of the candies in the chosen box.  His objective is as follows:    * Any two neighboring boxes contain at most x candies in total.  Find the minimum number of operations required to achieve the objective.

Submission attempts (in order):
N, x = map(int, input().split())
a = list(map(int, input().split()))
cnt = 0
for i in range(N-1):
    tmp = (a[i] + a[i+1]) - x
    if tmp > 0:
        continue
    cnt += tmp
    a[i] -= tmp - a[i+1]
    a[i+1] = 0
print(cnt)


N, x = map(int, input().split())
a = list(map(int, input().split()))
cnt = 0
for i in range(N-1):
    tmp = (a[i] + a[i+1]) - x
    cnt += tmp
    a[i] -= tmp - a[i+1]
    a[i+1] = 0
print(cnt)


N, x = map(int, input().split())
a = list(map(int, input().split()))
# N, x = 6, 1
# a = [1, 6, 1, 2, 0, 4]
# N, x = 5, 9
# a = [3, 1, 4, 1, 5]
# N, x = 2, 0
# a = [5, 5]

cnt = 0
for i in range(1, N):
    if a[i-1] + a[i] > x:
        rem = a[i-1] + a[i] - x
        cnt += rem
        a[i - 1] -= rem - a[i]
        a[i] = 0
print(a)
print(cnt)


N, x = map(int, input().split())
a = list(map(int, input().split()))
cnt = 0
for i in range(1, N):
    if a[i-1] + a[i] > x:
        rem = a[i-1] + a[i] - x
        cnt += rem
        if rem >= a[i]:
            a[i - 1] -= rem - a[i]
            a[i] = 0
        else:
            a[i] -= rem
print(cnt)



Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: There is a string s of length 3 or greater. No two neighboring characters in s are equal.  Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:    * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.  The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Submission attempts (in order):
s=input();l = len;p = print
if (s[0]==s[-1] * l(s)%2==0) + (s[0]!=s[-1] * l(s)%2==1):
    p("First")
else:
    p("Second")
s=input();l = len;p = print
if (s[0]==s[-1] * l(s)%2==0) + (s[0]!=s[-1] * l(s)%2==1):
    p("First")
else:
    p("Second")


s=input();l = len;p = print
if (s[0]==s[-1] and l(s)%2==0) or (s[0]!=s[-1] and l(s)%2==1):
    p("First")
else:
    p("Second")


Verdicts: Runtime Error

Accepted
Problem: There is a string s of length 3 or greater. No two neighboring characters in s are equal.  Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:    * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.  The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Submission attempts (in order):
S = input()
if (S[0] == S[-1]) ^ (len(S) % 2 == 0):
    ans = "First"
else:
    ans = "Second"
print(ans)


S = input()
ans = "First"
if (S[0] == S[-1]) ^ (len(S) % 2 == 0):
    ans = "Second"
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: There is a string s of length 3 or greater. No two neighboring characters in s are equal.  Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:    * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.  The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Submission attempts (in order):
s = input()
if (len(s)%2==0)^(s[0]==s[-2]):
  print('Second')
else:
  print('First')

s = input()
if (len(s)%2==0)^(s[0]==s[-1]):
  print('Second')
else:
  print('First')

Verdicts: Wrong Answer

Accepted
Problem: There is a string s of length 3 or greater. No two neighboring characters in s are equal.  Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:    * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.  The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Submission attempts (in order):
s = input()
if (s[0] == s[1])^(len(s)%2 == 1):
  print("First")
else:
  print("Second")

s = input()
t = ("First","Second")
print(t[(len(s)%2==0)^(s[0]==s[-1])])

Verdicts: Wrong Answer

Accepted
Problem: There is a string s of length 3 or greater. No two neighboring characters in s are equal.  Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:    * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.  The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

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

if (s[0] == s[-1]) ^ (len(s) & 1):
    print('Second')
else:
    print('First')

s = input()

if (s[0] == s[-1]) ^ (len(s) & 1):
    print('First')
else:
    print('Second')


Verdicts: Wrong Answer

Accepted
Problem: There is a string s of length 3 or greater. No two neighboring characters in s are equal.  Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:    * Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.  The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Submission attempts (in order):
#!/usr/bin/env python
# -*- coding:utf-8 -*-

s = input()
if len(s) % 2 == 0 ^ s[0] == s[-1]:
    print("Second")
else:
    print("First")


#!/usr/bin/env python
# -*- coding:utf-8 -*-

s = input()
if (len(s) % 2 == 0) == (s[0] == s[-1]):
    print("Second")
else:
    print("First")


#!/usr/bin/env python
# -*- coding:utf-8 -*-

s = input()
if (len(s) % 2 == 0) and (s[0] == s[-1]):
    print("First")
elif (len(s) % 2 == 0) or (s[0] == s[-1]):
    print("Second")
else:
    print("First")


Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
s = input()
while "BB" in s or "WW" in s:
	s.replace("BB", "B")
	s.replace("WW", "W")

print(len(s)-1)

s = input()
ans = -1
b = ""
for l in s:
	if b != l:
		ans += 1
	b = l
print(ans)

Verdicts: Time Limit Exceeded

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
S = input()
print(sum([a!=b for a,b in zip(S[:-1],S[1:])])

S = input()
print(sum([a!=b for a,b in zip(S[:-1],S[1:])]))

Verdicts: Runtime Error

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
s=input()
ans = 0
for I in range(len(s)-1):
 if s[i]!=s[i+1]:
  ans += 1
print(ans)

s=input()
ans = 0
for i in range(len(s)-1):
 if s[i]!=s[i+1]:
  ans += 1
print(ans)


Verdicts: Runtime Error

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
s = input()
prev = s[0]
ans = 0

for c in s:
    if c != prev:
        ans += 1
        prev = c
print(ans-1)


s = open('input', 'r').read()
prev = s[0]
ans = 0

for c in s:
    if c != prev:
        ans += 1
        prev = c
print(ans-1)


s = input()
prev = s[0]
ans = 0

for c in s:
    if c != prev:
        ans += 1
        prev = c
print(ans)


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
s=input()
cnt=0
for i range(1,len(s)):
  if s[i]!=s[i-1]:
    cnt+=1
print(cnt)

s=input()
cnt=0
for i in range(1,len(s)):
  if s[i]!=s[i-1]:
    cnt+=1
print(cnt)

Verdicts: Runtime Error

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):

s_long = input()

count = 0
prev = ''

for s in s_long:
    if prev == ''
        prev = s
        continue

    if prev != s:
        count += 1
        prev = s

print(count)

s_long = input()

count = 0
prev = ''

for s in s_long:
    if prev == '':
        prev = s
        continue

    if prev != s:
        count += 1
        prev = s

print(count)


Verdicts: Runtime Error

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
print('AAA')



import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  

print('AAA')



import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  


import  numpy as np
print('hoge!')
  

print('AAA')



import  numpy as np
print('hoge!')
  

from itertools import groupby
print(sum(1 for _ in groupby(input())) - 1)


Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Submission attempts (in order):
S=input();
DifF=0;
DifE=0;
First=S[0];
len=len(S)
End=S[len-1];
if First==End:
    for i in range(len):
    	if S[i]!=First:
    		DifF+=1;
    		DifE+=1;
elif First!=End:
    for i in range(len):
    	if S[i]!=First:
    		DifF+=1;
    	elif S[i]==First:
    	    DifE+=1;
if DifE >= DifF:
    print(DifF);
else:
    print(DifE);

S='WBWBWBWBWBWB';
DifF=0;
First=S[0];
len=len(S)
End=S[len-1];
for i in range(len-1):
    if S[i]!=S[i+1]:
        DifF+=1;
print(DifF);

S=input();
DifF=0;
First=S[0];
len=len(S)
End=S[len-1];
for i in range(len-1):
    if S[i]!=S[i+1]:
        DifF+=1;
print(DifF);

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

Submission attempts (in order):
n = int(input())
s = str(input().split())
t = str(input().split())
a = set(s+t)
print(a)
print(len(a)-3)

n = int(input())
s = str(input())
t = str(input())
for i in range (n+1):
  b = s[0:i] + t[0:n]
  if b[0:n] == s:
    n += i
    break
print(n)

Verdicts: Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

Submission attempts (in order):
n=int(input())
A=input()
B=input()
s=0

ans=2*n
for i in range(n)
  if A[-i-1:]!=B[:i+1]:
    s=1
    print(ans)
    exit()
  else:
    ans-=1
print(ans)

n=int(input())
A=input()
B=input()
s=0

ans=n
D=[i for i in range(n)][::-1]
for i in D:
  if A[-i-1:]==B[:i+1]:
    
    print(ans)
    exit()
  else:
    ans+=1
print(ans)

Verdicts: Runtime Error

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

Submission attempts (in order):
N = int(input())
s = input()
t = input()
num = 0
for i in range(0, N):
    if t[:i] == s[N-i-1:]:
        num = i
print(2*N - num)

N = int(input())
s = input()
t = input()
num = 0
for i in range(N):
    if t[:i+1] == s[-i-1:]:
        num = i+1
print(2*N - num)
        


Verdicts: Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

Submission attempts (in order):
N = int(input())
s = input()
t = input()
for i in range(0,N):
    if s[-i-1] !=t[i]:
        break
if s == t:
    print(s)
else:
    print(s + t[i:])


N = int(input())
s = input()
t = input()
mx = 0
for i in range(0,N):
    cnt = 0
    for j in range(0,N):
        if i+j >= N: break
        if s[i+j] == t[j]:
            cnt += 1
    mx = max(mx,cnt)
print(len(s+t[mx:]))


Verdicts: Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

Submission attempts (in order):
import sys


n=int(input())
s=input()
t=input()
test=s+t
if(s==t):
    print(s)
    sys.exit()
for i in range(1,n+1)[::-1]:
    test=s[0:-i]+t
    if(test[:n]==s and test[-n:]==t):
        print(test)
        sys.exit()
        
print(s+t)

import sys


n=int(input())
s=input()
t=input()
test=s+t
if(s==t):
    print(len(s))
    sys.exit()
for i in range(1,n+1)[::-1]:
    test=s[0:-i]+t
    if(test[:n]==s and test[-n:]==t):
        print(len(test))
        sys.exit()
        
print(len(s+t))

Verdicts: Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

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

f = 1
for i in range(n):
    if s[i:] == t[:n-i]:
        print(s + t[n-i:])
        f = 0
        break
if f:
    print(s + t)

n = int(input())
s = input()
t = input()
cnt = 0
for i in range(n):
    if s[n-i-1:] == t[:i+1]:
        cnt = len(t[:i+1])
print(2*n-cnt)

Verdicts: Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

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

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

print(n * 2 - i)

n = int(input())
s = input()
t = input()
cnt = 0
for i in range(n):
  if s[-1-i:] == t[:i+1]:
    cnt = i + 1
print(n*2-cnt)

Verdicts: Wrong Answer

Accepted
Problem: Snuke is interested in strings that satisfy the following conditions:    * The length of the string is at least N.   * The first N characters equal to the string s.   * The last N characters equal to the string t.  Find the length of the shortest string that satisfies the conditions.

Submission attempts (in order):
# coding: utf-8
N, s, t = int(input()), input(), input()

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

N = int(input())
s = input()
t = input()

t_ = ""
for i in range(N+1):
    if s[i:] == t[:N-i]:
        char = s + t[N-i:]
    else:
        char = s + t
print(len(char))

N = int(input())
s = input()
t = input()

for i in range(N+1):
    if s[i:N] == t[:N-i]:
        print(N + i)
        break

Verdicts: Wrong Answer

Wrong Answer

Accepted