Description
Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two
pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
Input
The first input line contains integer n (1?≤?n?≤?105) — amount of squares in the stripe. The second line contains
n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
Output
Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don‘t forget that it‘s allowed to cut the stripe along the squares‘ borders only.
Sample Input
Input
91 5 -6 7 9 -16 0 -2 2
Output
3
Input
31 1 1
Output
0
Input
20 0
Output
1
题意:
一张纸上画着n个格子，每个格子上写着一个整数（可能为负整数），把这些格子分成两段，使得这两段之和相等，求出一共有多少种方案。
思路：
在输入n个数的时候，用一个数组保存前i段之和，输入完毕之后，求出其总和，再用一个循环，看前i段之和是不是总和的两倍，如果是，那么ans++，循环完毕之后，ans就是答案。要注意的是要用long long 保存，否则会数据溢出。
代码：
#include<cstdio>
int a[100010];
long long b[100010];
using namespace std;
int main()
{
long long ans;
long long sum;
int n;
while(scanf("%d",&n)!=EOF)
{
ans=0;
sum=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
b[i]=sum;
// printf("b[%d]%d\n",i,b[i]);
}
// printf("%d\n",sum);
for(int i=1;i<n;i++)
if(2*b[i]==sum)
ans++;
printf("%I64d\n",ans);
}
return 0;
}
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