HDU 4334 trouble

Trouble

Time Limit: 10000/5000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 4375 accepted submission (s): 1297

Problem descriptionhassan 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?

 

Inputfirst 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.

 

Outputfor 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 input221-11-11-11-11-131 2 3-1-2-34 5 6-1 3 2-4-10-1

 

Sample outputnoyes

 1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #include <string.h> 5 #include <set> 6 using namespace std; 7 #define ll long long 8 ll a[5][300],s1[90000],s2[90000],s3[90000],sn1,sn2,sn3; 9 int main()10 {11     int n,t,i,j,k;12     cin>>t;13     while(t--)14     {15         cin>>n;16         for(i=0; i<5; i++)17             for(j=0; j<n; j++)18                 scanf("%I64d",&a[i][j]);19         sn1=sn2=sn3=0;20         for(i=0; i<n; i++)21             for(j=0; j<n; j++)22                 s1[sn1++]=a[0][i]+a[1][j];23         for(i=0; i<n; i++)24             for(j=0; j<n; j++)25                 s2[sn2++]=a[2][i]+a[3][j];26         for(i=0; i<n; i++)27             s3[sn3++]=-a[4][i];28         sort(s1,s1+sn1);29         sort(s2,s2+sn2);30         sort(s3,s3+sn3);31         int ok=0;32         for(i=0;!ok&&i<sn3;i++)33         {34             j=0,k=sn2-1;35             while(j<sn1&&k>=0)36             {37                 if(s1[j]+s2[k]==s3[i])38                 {39                     ok=1;40                     break;41                 }42                 while(j<sn1&&k>=0&&s1[j]+s2[k]>s3[i])k--;43                 while(j<sn1&&k>=0&&s1[j]+s2[k]<s3[i])j++;44             }45         }46         if(ok)printf("Yes\n");47         else printf("No\n");48     }49 }

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