二分基础

<span style="color:#3333ff;">/*A - 二分法 基础
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).
By Grant Yuan
2014.7.14
二分
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
long long a[4002],b[4002],c[4002],d[4002];
int n;
long long num1[16000004];
long long num2[16000004];
int top;
long long sum;
int binary(long k,int left,int right)
{
     int i;
    while(left<=right){
      int mid=(left+right)/2;
      int num=0;
           if(num2[mid]==k)
       {
         num=1;
         for(i=mid-1;i>=0&&num2[i]==k;i--)  num++;
         for(i=mid+1;i<n*n&&num2[i]==k;i++)  num++;
         return num;
      }
     else if(num2[mid]>k)
       right=mid-1;
      else left=mid+1;
   }
   return 0;

}

int main()
{
    cin>>n;
    long long flag;
    for(int i=0;i<n;i++)
       cin>>a[i]>>b[i]>>c[i]>>d[i];
    top=-1;
    sum=0;
    for(int i=0;i<n;i++)
      for(int j=0;j<n;j++)
     {
         num1[++top]=a[i]+b[j];
         num2[top]=c[i]+d[j];
     }
     sort(num1,num1+top+1);
     sort(num2,num2+top+1);
     for(int i=0;i<=top;i++)
       {
           flag=-num1[i];
           sum+=binary(flag,0,top+1);
       }
       cout<<sum<<endl;
       return 0;
}
</span>