Contest1376 - "师创杯"烟台大学第二届ACM程序设计精英赛复现Problem H: H-Sum 3s

Problem H: H-Sum 3s

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 77 Solved: 10

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Description

You are given a number sequence a1,a2,a3...,an , your task is to find if there is a pair of interger (i,j) that ai+a(i+1)+..+aj equals to 0 and i<=j;

Input

Input consists of multiple test cases. For each case, the first line of input contains an integer n, the next line follows n integers. (n>=1 && n<=10^5 |ai|<=10^4)

Output

For each case, if there is at least one pair of integer (i,j) meet the requirement, print “YES”, otherwise print “NO” .

Sample Input

51 2 3 4 553 4 -2 -3 1

Sample Output

NOYES

HINT

AC代码：

#include <iostream>
using namespace std;
int main(){
    int n;
    while(cin>>n){
        int a
;
        int sum=0,i=0,k=0,s1,t1,t2,j;;
        while(i<n){
            cin>>a[i];
            sum+=a[i];
            if(a[i]==0||sum==0)k++;
            if(k==0){
                for(j=0,t2=0;j<i-1;j++){
                    t1=a[j];
                    t2+=t1;
                    s1=sum-t2;
                    if(s1==0){
                        k++;
                        break;
                        }
                    }
                }
            i++;
        }
        if(k==0) cout<<"NO"<<'\12';
        else cout<<"YES"<<'\12';
    }
    return 0;
}

运行结果：



心得：

这看起来是一道很简单的题，2个循环就出来了。但如果这样oj平台会给出时间超限的提示。所以要想办法减少计算时间，我是在输入数据的同时完成部分运算。谢天谢地在错误9次后终于ac了。感觉棒！