stripe2

桃扇骨 2022-05-30 10:56 132阅读 0赞

C. Stripe 2

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

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 three pieces so that the sum of numbers from each piece is equal to the sum of numbers from any 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 three non-empty pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

Examples

input

Copy

4
    1 2 3 3

output

input

Copy

5
    1 2 3 4 5

output

这个题是个水题， 只要统计分割点之间的匹配关系就可以， 但是自己就是做不对，因为自己分割点弄混了， 导致接二连三的出错，最后答案可能是long long 以为1e5 全是零就超INT了。

#include<cstdio>
    #include<algorithm>
    using namespace std;
    
    const int maxn=1e5+10;
    int num[maxn], sum[maxn], tt1[maxn], cnt1, tt2[maxn], cnt2;
    
    int main(){
        int n;
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d", num+i);
            sum[i]=sum[i-1]+num[i];
        }
        if(sum[n]%3 != 0)
        {
            printf("0\n");
            return 0;
        }
    
        int stand=sum[n]/3;
    
        for(int i=1; i<n; i++)
        {
            if(sum[i]==stand)
            {
                tt1[++cnt1]=i+1;
            }
        }
        for(int i=1; i<n; i++)
        {
            if(sum[i]==stand*2)
            {
                tt2[++cnt2]=i+1;
            }
        }
    
        long long ans=0;
        for(int i=1; i<=cnt1; i++)
        {
            int *p=upper_bound(tt2+1, tt2+1+cnt2, tt1[i]);
    
            ans+=(tt2+cnt2-p+1);
        }
        printf("%I64d\n", ans);
        return 0;
    }