The SUM problem can be formulated as follows: given four lists A;B;C;D of integer values, compute
how many quadruplet (a; b; c; d) 2 AB C D are such that a+b+c+d = 0. In the following, we
assume that all lists have the same size n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.
The rst line of the input le contains the size of the lists n (this value can be as large as 4000).
We then have n lines containing four integer values (with absolute value as large as 228) that belong
respectively to A;B;C and D.
Output
For each test case, your program has to write the number quadruplets whose sum is zero.
The outputs of two consecutive cases will be separated by a blank line.

 #include<cstdio>
 #include<cstring>
 int abs(int x)
 {
     ) return x;
     return -x;
 }
 ;
 ],b[],c[],d[],first[],next[],num[];
 int main()
 {
     int i,j,k,n,p,q,x,y,z,t,ans;
     scanf("%d",&t);
     while (t--)
     {
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         memset(c,,sizeof(c));
         memset(d,,sizeof(d));
         memset(first,,sizeof(first));
         memset(next,,sizeof(next));
         memset(num,,sizeof(num));
         ans=;
         scanf("%d",&n);
         ;i<=n;i++)
           scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
         ;i<=n;i++)
           ;j<=n;j++)
           {
               x=a[i]+b[j];
               p=abs(x%m);
               next[(i-)*n+j]=first[p];
               first[p]=(i-)*n+j;
               num[(i-)*n+j]=x;
           }
         ;i<=n;i++)
           ;j<=n;j++)
           {
               x=-c[i]-d[j];
               p=abs(x%m);
               for (k=first[p];k;k=next[k])
                 if (x==num[k]) ans++;
           }
         printf("%d\n",ans);
         if (t) printf("\n");
     }
 }

枚举a+b,把所有值存起来,然后枚举-c-d,在a+b中查找。

具体查找方法是哈希,除k取余法即可。

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