A. Spit Problem
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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.
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 xiand 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.
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
2
0 1
1 -1
YES
3
0 1
1 1
2 -2
NO
5
2 -10
3 10
0 5
5 -5
10 1
YES
/* ***********************************************
Author :
Created Time :2015/6/14 0:27:17
File Name :7.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1<<30
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
bool cmp(int a,int b){
return a>b;
}
int a[maxn];
int b[maxn];
int main()
{
#ifndef ONLINE_JUDGE
//freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int n;
while(cin>>n){
int mark=0;
for(int i=1;i<=n;i++){
cin>>a[i]>>b[i];
}
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
if(a[i]+b[i]==a[j]&&a[j]+b[j]==a[i]){
mark=1;break;
}
}
if(mark)break;
}
if(mark)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
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