HDU 4334

Trouble

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2961 Accepted Submission(s): 893

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

 1 //大致题意：判断五个数之和是否可能为0 

 2 #include <iostream>

 3 #include <cstring>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 typedef __int64 LL;//用define时出了错误 

 8 LL a[210];

 9 LL b[210];

10 LL c[210];

11 LL f[202*202];

12 LL g[202*202];

13 int num;

14 

15 int main()

16 {

17     int i,j,k,T;

18     cin>>T;

19     while(T --)

20     {

21         cin>>num;

22         //memset(f, 0, sizeof f);

23         //memset(g, 0, sizeof g);

24         int cnt1 = 0;

25         int cnt2 = 0;

26         for(i = 0; i < num; i ++) 

27           cin>>a[i];

28         for(i = 0; i < num; i ++) 

29           cin>>b[i];

30         for(i = 0; i < num; i ++)

31             for(j = 0; j < num; j ++)

32                 f[cnt1++] = a[i] + b[j];

33         for(i = 0; i < num; i ++) 

34           cin>>a[i];

35         for(i = 0; i < num; i ++) 

36           cin>>b[i];

37         for(i = 0; i < num; i ++)

38             for(j = 0; j < num; j ++)

39                 g[cnt2++] = a[i] + b[j];

40         for(i = 0; i < num; i ++) 

41           cin>>c[i];

42         sort(f, f + cnt1);

43         sort(g, g + cnt2);

44         sort(c, c + num);

45         bool flag = 0;

46         for(i = 0; i < num; i ++)

47         {

48             LL temp = c[i];

49             j = 0;

50             k = cnt2 - 1;

51             while(j < cnt1 && k >= 0)

52             {

53                 if(f[j] + g[k] == (-temp))

54                 {

55                     flag = 1;

56                     break;

57                 }

58                 if(f[j] + g[k] <(-temp)) 

59                     j++;

60                 else 

61                     k--;

62             }

63         }

64         if(flag)

65           cout<<"Yes"<<endl;

66         else

67           cout<<"No"<<endl;

68     }

69     return 0;

70 }