BestCoder Round #61 (div.2) HDU5522 Numbers

最新推荐文章于 2017-08-19 11:53:20 发布

原创最新推荐文章于 2017-08-19 11:53:20 发布 · 327 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #HDU #BestCoder #模拟

ACM 同时被 3 个专栏收录

121 篇文章

订阅专栏

模拟

53 篇文章

订阅专栏

HDU

49 篇文章

订阅专栏

本文探讨了排序算法和查找算法的优化策略，通过实例展示了如何提高算法效率，包括使用预排序、二分查找和哈希表等技术，旨在帮助开发者解决实际问题时提升程序性能。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/262144 K (Java/Others)
Total Submission(s): 95 Accepted Submission(s): 72

Problem Description

There are n numbers

A1,A2....An ,your task is to check whether there exists there different positive integers i, j, k (

1≤i,j,k≤n ) such that

Ai−Aj=Ak

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

Sample Output

  
   YES
NO
YES

Source

BestCoder Round #61 (div.2)

出题人：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.

#include
  
   
#include
   
    
#include
    
     
#include
     
      
using namespace std;
int a[105];
int b[1001];
int main()
{
    int n,i,j,t,flag;
    while(~scanf("%d",&n))
    {
        flag=0;
        for(i=0; i<=1000; i++)
            b[i]=0;
        for(i=0; i
      
       =2; j--)
        {
            for(i=j-1; i>=1; i--)
            {
                int temp=a[j]-a[i];
                if(temp==a[i]&&b[temp]>=2)
                {
                    flag=1;
                    break;
                }
                if(temp!=a[i]&&b[temp]>=1)
                {
                    flag=1;
                    break;
                }
            }
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}