UVA1152 4 Values whose Sum is 0, uva11524values

UVA1152 4 Values whose Sum is 0, uva11524values

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) $ \ in $ AxBxCxD 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 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 test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the number quadruplets whose sum is zero.
Sample Input
1

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

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 ).

 

Problem: it is best not to time out .. Enumerate a and B first, and then check whether the value of-(c + d) is binary optimization.

AC code:

#include <algorithm>#include <iostream>using namespace std;const int Max = 4000 + 10;int a[Max],b[Max],c[Max],d[Max];int ab[17000000]；int total;int main(){    int t;    cin>>t;    while(t--)    {        int n;        cin>>n;        for(int i=0; i<n; i++)        {            cin>>a[i]>>b[i]>>c[i]>>d[i];        }        int k=0;        for(int i=0;i<n; i++)        {            for(int j=0;j<n; j++)            {                ab[k]=a[i]+b[j];                k++;            }        }        sort(ab,ab+k);        total=0;        int s,l,r,mid;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                int x=-c[i]-d[j];                l=0,r=k-1;                while(l<=r)                {                    mid=(l+r)/2;                    if(ab[mid]>x)                        r=mid-1;                    else if(ab[mid]<x)                        l=mid+1;                    else                    {                        for(s=mid;s>=0;s--)                        {                            if(ab[s]==x)                                total++;                            else                               break;                        }                        for(s=mid+1; s<k; s++)                        {                            if(ab[s]==x)                                total++;                            else                               break;                        }                        break;                    }                }            }        }        cout<<total<<endl;        if(t>0)            cout<<endl;    }    return 0;}