4 Values whose Sum is 0

 

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

Submit  Status

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

5

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\n",cout);

56     }

57     return 0;

58 }