zoj 3672 Gao The Sequence

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4915题意：a[k]-一个任意的数，这个数要等于a[1]~a[k]每个数减去任意一个数，经过多次这样的变换到达目标b序列，能到达就yes不能到达距no思路：既然a[k]前面数相减等于这个数，则所有减去的数之和sum一定数偶数。其中最大的减数的2倍一定要小于这个sum，如果不满足就不可能从a序列到b序列。Gao The Sequence

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

#include <stdio.h>  
#include <string.h>  
#include <algorithm>  
using namespace std;  
int max(int x,int y)
{
	if(x>y)
		return x;
	else
		return y;
}
int  main()
{
    long long maxn,a[10000],b[10000],sum,q;
	int t,i;
    while(~scanf("%d",&t))
    {
           sum=maxn=0;
        for(i=0;i<t;i++)
          {
             scanf("%lld%lld",&a[i],&b[i]);
              q=a[i]-b[i];
              sum+=q;
              maxn=max(maxn,q);
          }
          if(sum%2!=0)
              printf("NO
");
          else if(2*maxn>sum)
               printf("NO
");
            else
            printf("YES
");


    }
    return 0;
}
/*
6
10 1
5 3
16 5
9 3
9 2
14 1
*/