CodeForces - 29A - Spit Problem

60 阅读 0 评论 40 点赞

我是靠谱客的博主魔幻羽毛，最近开发中收集的这篇文章主要介绍CodeForces - 29A - Spit Problem，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

题目出处：codeforces - 29A

In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.

The trajectory of a camel's spit is an arc, i.e. if the camel in position x spits d meters right, he can hit only the camel in position x + d, if such a camel exists.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the amount of camels in the zoo. Each of the following n lines contains two integers xi and di ( - 104 ≤ xi ≤ 104, 1 ≤ |di| ≤ 2·104) — records in Bob's notepad. xi is a position of the i-th camel, and di is a distance at which the i-th camel spitted. Positive values of di correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.

Output

If there are two camels, which spitted at each other, output YES. Otherwise, output NO.

        Sample test(s) 
      
          input 
        
2
0 1
1 -1

          output 
        
YES

          input 
        
3
0 1
1 1
2 -2

          output 
        
NO

          input 
        
5
2 -10
3 10
0 5
5 -5
10 1

          output 
        
YES

水题，计算下每个终点，比较起点和终点不在同一点，就吐不到

the accepted code like this :

//
//  main.cpp
//  CF29A
//
//  Created by IntelliegeWither on 15/12/2.
//  Copyright © 2015年 IntelliegeWither. All rights reserved.
//

#include<stdio.h>
int main()
{
    int a[105],d[105],zd[105];
    int n;
    int i,j,sign=0;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d%d",&a[i],&d[i]);
        zd[i]=a[i]+d[i];
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(i!=j&&a[i]==zd[j]&&a[j]==zd[i])//相互吐不中，标记
            {
                sign=1;
            }
        }
    }
    if(sign)
    {
        printf("YESn");
    }
    else
    {
        printf("NOn");
    }
}