HDU4334(多重循环)

Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3459    Accepted Submission(s): 1091

Problem Description

Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?

 

Input

First line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.

 

Output

For each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".

 

Sample Input

   
   
   
   

    
    
    
    2
2
1 -1
1 -1
1 -1
1 -1
1 -1
3
1 2 3
-1 -2 -3
4 5 6
-1 3 2
-4 -10 -1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    No
Yes
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    多重循环，也能过，注意优化数组的访问方式
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 ll;

ll da1[201],da2[201],da3[201];
ll two[201*201];
ll four[201*201];

int main()
{
	int i,j,k1,t,n,k2;
	bool flag=false;
	cin>>t;
	while(t--)
	{
		scanf("%d",&n);
		for(j=0;j<n;j++)
		{
			scanf("%I64d",da1+j);
		}
		for(j=0;j<n;j++)
		{
			scanf("%I64d",da2+j);
		}

		k1=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				two[k1++]=da1[i]+da2[j];
			}
		}
		sort(two,two+k1);


		for(j=0;j<n;j++)
		{
			scanf("%I64d",da1+j);
		}
		for(j=0;j<n;j++)
		{
			scanf("%I64d",da2+j);
		}
		k2=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				four[k2++]=da1[i]+da2[j];
			}
		}
		sort(four,four+k2);

		for(j=0;j<n;j++)
		{
			scanf("%I64d",da1+j);
		}
		sort(da1,da1+n);

		flag=false;
		for(i=0;i<n;i++)
		{
			int left=0;
			int right=k2-1;
			while(left<k1&&right>=0)
			{
				if(-da1[i]==four[right]+two[left])
				{
					flag=true;
					break;
				}
				if(four[right]+two[left]<-da1[i])
					left++;
				else 
					right--;
			}
			if(flag)
				break;
		}
	  if(flag)
          puts("Yes");
        else
          puts("No");

	}
	return 0;
}