Codeforces 18C C. Stripe

Codeforces 18C  C. Stripe

链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/E

题目：

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

9 1 5 -6 7 9 -16 0 -2 2

Output

3

Input

3 1 1 1

Output

0

Input

2 0 0

Output

1

题意：

求能将前面和后面分为和相等的两部分的次数

分析：

1.求前K个数的和b[k]和后n-k+1个数的和c[k+1]

2.比较大小，若相等记一次

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int maxn=100001;

int a[maxn],b[maxn],c[maxn]; //数组很大时要放在外面，不然会出现WA
    
int main()
{
    int n;
    scanf("%d",&n);
    b[0]=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        b[i]=b[i-1]+a[i];    //求前i项的和
    }
    for(int j=n;j>=1;j--)
    {
        c[j]=c[j+1]+a[j];   //求后n-j+1项的和
    }
    int ans=0;
    for(int k=1;k<n;k++)
    {
        if(b[k]==c[k+1])  //比较
            ans++;
    }
    printf("%d
",ans);
    return 0;
}

读了好久的题目，一开始对负数那里不理解，仔细分析了一下，又听别人说了就理解了。