[hdu P4334] Trouble

[hdu P4334] Trouble

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?

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 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

第一次手写Hash表。。。只不过跑的有点慢。。。

发现hash最常用的还是用邻接表挂链。。其实还蛮好写的，只是以前没写过而已。

code：

（第一次写class）

 %:pragma GCC optimize()
 #include<bits/stdc++.h>
 #define LL long long
 #define r register
 #define Ms(a,x) memset(a,x,sizeof a)
 using namespace std;
 ;
 ][N],sum;
 class Hashmap {
     private:
         #define p 2040803
         LL lnk[p+],son[N*N]; int nxt[N*N],tot;
     public:
         ),Ms(nxt,-),tot=;}
         void insert(LL x) {
             LL k=x%p; ) k+=p;
             ) {lnk[k]=++tot,son[tot]=x; return;}
             for (r int j=lnk[k]; ; j=nxt[j])
                 ) {nxt[j]=++tot,son[j]=x; return;}
         }
         bool find(LL x) {
             LL k=x%p; ) k+=p;
             ;
             ;
         }
 }app;
 inline LL read() {
     LL x=,f=; char ch=getchar();
     :,ch=getchar();
     +ch-',ch=getchar();
     return x*f;
 }
 int main() {
     for (int T=read(); T; T--) {
         n=read(),app.clear();
         ; i<=; i++)
             ; j<=n; j++) a[i][j]=read();
         ; i<=n; i++)
             ; j<=n; j++)
                 sum=-(a[][i]+a[][j]),app.insert(sum);
         ;
         ; i<=n&&!tag; i++)
             ; j<=n&&!tag; j++)
                 ; k<=n&&!tag; k++) {
                     sum=a[][i]+a[][j]+a[][k];
                     ;
                 }
         printf("%s\n",tag?"Yes":"No");
     }
     ;
 }