二分查找的应用

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 using namespace std;
 5 
 6 int a1[4001],a2[4001];
 7 int b1[4001],b2[4001];
 8 int sum1[4001*4000],sum2[4001*4000];
 9 
10 int  binarySearch(int left,int right,int key)
11 {
12     __int64 ans=0; //64位整数
13     while(left<=right)
14     {
15         int mid = (left + right) / 2;
16         if(sum2[mid] == key)
17         {
18             ans++;
19             for(int i=mid-1; i>=left; i--)   //搜到后可能相邻的左边有相等的也满足条件的数！
20             {
21                 if(sum2[i]!=key)
22                 {
23                     break;
24                 }
25                 ans++;
26             }
27             for(int j=mid+1; j<=right; j++)
28             {
29                 if(sum2[j]!=key)
30                 {
31                     break;
32                 }
33                 ans++;
34             }
35             return ans;
36         }
37         else if(sum2[mid] < key) left = mid+1;
38         else right = mid -1;
39     }
40     return 0;
41 }
42 
43 int main()
44 {
45     int n;
46     cin>>n;
47     __int64 ans=0;
48     for(int i = 0; i < n; i++)
49     {
50         cin>>a1[i]>>a2[i]>>b1[i]>>b2[i];
51 
52     }
53     int times = 0;
54     for(int i = 0; i < n; i++) //四数相加转化为两数，通过两两随机相加，以便二分
55         for(int j = 0; j < n; j++)
56         {
57             sum1[times] = a1[i] + a2[j];
58             sum2[times] = b1[i] + b2[j];
59             times++; //加时的下表
60         }
61     sort(sum2,sum2 + times);
62 
63     for(int i = 0; i < times; i++)
64     {
65         int key = -sum1[i];
66         ans += binarySearch(0,times-1,key);
67     }
68     cout<<ans<<endl;
69     return 0;
70 }