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?
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
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
View Code
1 #include <iostream> 2 #include <set> 3 #include <algorithm> 4 using namespace std; 5 6 __int64 a[5][205]; 7 __int64 b[2][40001]; 8 9 int main() 10 { 11 int T; 12 scanf("%d",&T); 13 while (T--) 14 { 15 int i,j; 16 int n,k=0; 17 scanf("%d",&n); 18 for (i=0;i<5;i++) 19 for (j=0;j<n;j++) 20 scanf("%I64d",&a[i][j]); 21 sort(a[4],a[4]+n); 22 int f=0; 23 for (i=0;i<n;i++) 24 for (j=0;j<n;j++) 25 b[0][f++]=a[0][i]+a[1][j]; 26 for (i=0;i<n;i++) 27 for (j=0;j<n;j++) 28 b[1][k++]=(a[2][i]+a[3][j]); 29 sort(b[0],b[0]+f); 30 sort(b[1],b[1]+k); 31 int p; 32 bool flag=0; 33 for (i=0;i<n;i++) 34 { 35 for (j=k-1,p=0;j>=0&&p<f;) 36 { 37 if (a[4][i]+b[1][j]+b[0][p]>0) 38 j--; 39 else 40 if (a[4][i]+b[1][j]+b[0][p]<0) 41 p++; 42 else 43 { 44 flag=1; 45 break; 46 } 47 } 48 } 49 if (flag) 50 printf("Yes\n"); 51 else printf("No\n"); 52 } 53 return 0; 54 }