多校4题目之Trouble

Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1870    Accepted Submission(s): 588

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

 

 

Source

2012 Multi-University Training Contest 4

 

 

Recommend

zhoujiaqi2010

这道题目确实比较坑爹 用set写了个代码现在还是TLE 时间卡得好死 开始以为是Dp 而且我DP也不熟练

写了很久算了一下时间复杂度应该能卡过去 不知道为什么却一直还是TLE

纠结了差不多三个小时 

其他的题目又不会做  总的来说还是能力的问题吧 我们这只队伍还是得全部提高  我也得提高 呵呵

题目意思：

给你五个序列 要你在每一个序列用找一个数 如果存在和为0的情况 就输出Yes 否则输出No 总共用120*10^6*log已经超过10^8 所以超时了 后来才算出来

害得我纠结了很久

解题报告是说开始用两个数组sum把分别计算s1+s2的和s3+s4的和

排序后   用指针下标i    指向sum1的开头 用指针下标J 指向sum2的结尾 如果

sum1+sum2+s[4][k]<0 i++

>0 j--

=0 直接跳出循环 说明已经找到了 

#include <iostream>
#include <set>
#include <algorithm>
using namespace std;

__int64 a[5][205];
__int64 b[2][40001];

int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        int i,j;
        int n,k=0;
        scanf("%d",&n);
        for (i=0;i<5;i++)        
            for (j=0;j<n;j++)            
                scanf("%I64d",&a[i][j]);                
            sort(a[4],a[4]+n);
            int f=0;
            for (i=0;i<n;i++)        
                for (j=0;j<n;j++)                                        
                    b[0][f++]=a[0][i]+a[1][j];            
                for (i=0;i<n;i++)        
                    for (j=0;j<n;j++)                                        
                        b[1][k++]=(a[2][i]+a[3][j]);    
                    sort(b[0],b[0]+f);
                    sort(b[1],b[1]+k);
                    int p;    
                    bool flag=0;                
                        for (i=0;i<n;i++)
                        {
                            for (j=k-1,p=0;j>=0&&p<f;)
                        {
                            if (a[4][i]+b[1][j]+b[0][p]>0)                                    
                                j--;                                    
                            else 
                                if (a[4][i]+b[1][j]+b[0][p]<0)
                                    p++;
                                else 
                                {
                                    flag=1;
                                    break;
                                }                                        
                        }
                    }
                    if (flag)                        
                        printf("Yes\n");                        
                    else printf("No\n");                    
    }
    return 0;
}