4thweek contest problem_E codeforces 18c前缀码之和问题

题目描述：

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?

分析：

1、输入第一个数字就存进数组为a[0]，后面每输入一个数字就将它加上前面的a[i-1]并存进a[i]。得到数组a

2、循环，从数组第一个开始循环找：左边a[i]==右边a[n-1]-a[i]。设置一个变量cnt记录这个条件成立时的分配次数。输出，此题解决。

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：

Input：

3 1 1 1

Output：

Input：

2 0 0

Output

1

代码及分析：

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 const int m=100050;
 5 long long a[m];
 6 
 7 int main()
 8 {
 9     int n,c,i,cnt=0,p=0;
10     scanf("%d",&n);
11     for(i=0;i<n;i++)
12     {
13         scanf("%d",&c);
14         a[i]=p+c;
15         p=a[i];
16 
17     }
18 
19     for(i=0;i<n-1;i++)
20        {
21            if(a[i]==(a[n-1]-a[i]))
22            cnt++;
23 
24        }
25      printf("%d
",cnt);
26 
27 }