HDU 5522 Numbers 暴力

Numbers

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5522

Description

给n个数A1,A2....An{A}_{1},{A}_{2}....{A}_{n}A​1​​,A​2​​....A​n​​，从中选3个位置不同的数A,B和C，问是否有一种情况满足A-B=C.

Input

There are multiple test cases, no more than 1000 cases.
First line of each case contains a single integer n.(3≤n≤100).
Next line contains n integers A1,A2....An.(0≤Ai≤1000)

Output

For each case output "YES" in a single line if you find such i, j, k, otherwise output "NO"

Sample Input

3
3 1 2
3
1 0 2
4
1 1 0 2

Sample Output

YES
NO
YES

HINT

题意

题解：

就n^2暴力，先排序然后从大到小枚举i，把右边的数用一个数组标记其出现过，再枚举左边的数判断其加上Ai是否出现过.

代码

#include<iostream>

#include<stdio.h>

#include<algorithm>

#include<map>

using namespace std;

map<int,int> H;

int a[];

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        H.clear();

        for(int i=;i<=n;i++)

            scanf("%d",&a[i]),H[a[i]]++;

        sort(a+,a++n);

        int flag = ;

        for(int i=;i<=n;i++)

        {

            H[a[i]]--;

            for(int j=;j<i;j++)

                if(H[a[j]+a[i]])

                    flag = ;

            if(flag)

                break;

        }

        if(flag)

            printf("YES\n");

        else

            printf("NO\n");

    }

}