POJ 2785 4 Values whose Sum is 0

Original question:
4 Values whose Sum is 0
Time limit:15000ms Memory limit:228000k
Total submissions:18834 accepted:5600
Case Time Limit:5000ms
Description

The SUM problem can formulated as Follows:given four lists A, B, C, D of an integer values, compute how many quadruplet ( A, B, C, D) ∈a x B x C x D is such that A + B + c + d = 0. In the following, we assume this all lists has 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 had 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 of the input file, your program have 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), (-3) 2, 22, 56,-46), (-32, 30,-75, 77), (-32,-54, 56, 30).
Source

Southwestern Europe 2005
Effect:
Give you a number of 6 and then give you 4 arrays of a,b,c,d. Each array contains 6 numbers, and now lets you find out how to find each one in A,b,c,d and 0. The number of output scenarios.

#include <bits/stdc++.h> #include <iostream> #include <algorithm> #include <vector> using
namespace Std;
FStream in,out;
Vector<int> a,b,c,d,s;
int n;
    int main () {Ios::sync_with_stdio (false);
    int a,b,c,d;
        while (cin>>n) {a.clear ();
        B.clear ();
        C.clear ();
        D.clear ();
        S.clear ();
            for (int i=0;i<n;i++) {cin>>a>>b>>c>>d;
            A.push_back (a);
            B.push_back (b);
            C.push_back (c);
        D.push_back (d);

        } for (int i=0;i<n;i++) for (int j=0;j<n;j++) S.push_back (C[i]+d[j]);
        Sort (S.begin (), S.end ());
        Long Long ans=0;
                for (int i=0;i<n;i++) {for (int j=0;j<n;j++) {int tmp=a[i]+b[j];
            Ans+=upper_bound (S.begin (), S.end (),-tmp)-lower_bound (S.begin (), S.end (),-tmp);
}
        }        cout<<ans<<endl;
} return 0;
 }

Answer:
In the Challenge Program design contest above see, very basic two-point search topic. C[i] and d[j] are stored in S. Then the S is sorted, enumerated a[i]+b[j], and then the binary finds the upper and lower bounds of S (A[i]+b[j])

Look at the discussion area This question also has the hash method, strives for later to fill up