4 Values whose Sum is 0

Time Limit:15000MS Memory Limit:228000KB 64bit IO Format:%I64d & %I64u

Description

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 ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .

Input

The first line of the input file 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 2 ²⁸ ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77

-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

 1 #include<cstdio>
 2 #include<string.h>
 3 #include<algorithm>
 4 #define MAXN 4400
 5 using namespace std;
 6 int A[MAXN],B[MAXN],C[MAXN],D[MAXN];
 7 int S[MAXN*MAXN];
 8 int lower_bound1(int low,int high,int num,int a[])
 9  {
10      int mid;
11      while(low<high)
12      {
13          mid=low+(high-low)/2;
14          if(a[mid]>=num) high=mid;
15          else low=mid+1;
16      }
17      return low;
18  }
19 int upper_bound1(int low,int high,int num,int a[])
20 {
21     int mid;
22     while(low<high)
23     {
24         mid=low+(high-low)/2;
25         if(a[mid]<=num) low=mid+1;
26         else
27             high=mid;
28     }
29     return low;
30 }
31 int main()
32 {
33     int n,i;
34     int p;
35     int cout=0;
36     int l,r,j;
37     while(scanf("%d",&n)!=EOF)
38     {
39         cout=0;
40         for(i=0;i<n;i++)
41             scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);
42        p=0;
43        for(i=0;i<n;i++)
44             for(j=0;j<n;j++)
45              S[p++]=A[i]+B[j];
46         sort(S,S+p);
47        for(i=0;i<n;i++)
48          for(j=0;j<n;j++)
49        {
50            int t=C[i]+D[j];
51            l=lower_bound1(0,p,-t,S);
52            r=upper_bound1(0,p,-t,S);
53             cout+=(r-l);
54        }
55        printf("%d
",cout);
56     }
57     return 0;
58 }