Poj 2785 4 values whose sum is 0 (half enumeration (bidirectional search ))

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 228 ) 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 -4626 -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).

 

AC code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <set> 7 using namespace std; 8 #define ll long long 9 #define N 400610 int n;11 int mp[N][N];12 int A[N],B[N],C[N],D[N];13 int CD[N*N];14 void solve(){15    for(int i=0;i<n;i++){16        for(int j=0;j<n;j++){17           CD[i*n+j]=C[i]+D[j];18        }19    }20    sort(CD,CD+n*n);21    ll ans=0;22    for(int i=0;i<n;i++){23        for(int j=0;j<n;j++){24           int cd = -(A[i]+B[j]);25           ans+=upper_bound(CD,CD+n*n,cd)-lower_bound(CD,CD+n*n,cd);26        }27    }28    printf("%I64d\n",ans);29 }30 int main()31 {32     while(scanf("%d",&n)==1){33        for(int i=0;i<n;i++){34           for(int j=0;j<4;j++){35              scanf("%d",&mp[i][j]);36           }37        }38        for(int i=0;i<n;i++){39           A[i] = mp[i][0];40           B[i] = mp[i][1];41           C[i] = mp[i][2];42           D[i] = mp[i][3];43        }44        solve();45     }46     return 0;47 }

 

Poj 2785 4 values whose sum is 0 (half enumeration (bidirectional search ))