zoj Gao The Sequence

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You are given a sequence of integers, A1,A2,...,An. And you are allowed a manipulation on the sequence to transform the origin sequence into another sequence B1,B2,...,Bn(Maybe the two sequences are same ). The manipulation is specified as the following three steps:

1.Select an integer Ai and choose an arbitrary positive integer delta as you like.

2.Select some integers Aj satisfying j < i, let‘s suppose the selected integers are Ak1,Ak2,...,Akt , then subtract an arbitrary positive integer Di from Aki (1 ≤ i ≤ t) as long as sum(Di) = delta.

3.Subtract delta from Ai.

The manipulation can be performed any times. Can you find a way to transform A1,A2,...,An to B1,B2,...,Bn ?

Input

The input consist of multiple cases. Cases are about 100 or so. For each case, the first line contains an integer N(1 ≤ N ≤ 10000) indicating the number of the sequence. Then followed by N lines, ith line contains two integers Ai and Bi (0 ≤ Bi ≤ Ai ≤ 4294967296).

Output

Output a single line per case. Print "YES" if there is a certain way to transform Sequence A into Sequence B. Print "NO" if not.

Sample Input

3
3 2
4 2
5 2
3
2 0
7 1
3 1

Sample Output

YES
NO

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long LL;
 8 
 9 LL a[10002];
10 int main()
11 {
12     LL n,i,x,y;
13     while(scanf("%lld",&n)>0)
14     {
15         LL max1=-1;
16         for(i=1;i<=n;i++)
17         {
18             scanf("%lld%lld",&x,&y);
19             a[i]=x-y;
20         }
21         sort(a+1,a+1+n);
22         max1=a[n];
23         LL sum=0;
24         LL hxl=a[n];
25         for(i=1;i<=n-1;i++)
26         {
27             sum=sum+a[i];
28             hxl=(hxl+a[i])%2;
29         }
30         if(max1>sum || hxl==1) printf("NO\n");
31         else printf("YES\n");
32     }
33     return 0;
34 }

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zoj Gao The Sequence

标签：style   blog   class   code   java   c   

原文地址：http://www.cnblogs.com/tom987690183/p/3724303.html