HRBEU 1121 二分枚舉

題目：
Kth Largest
TimeLimit: 1 Second   MemoryLimit: 32 Megabyte
Totalsubmit: 442   Accepted: 161  
Description
There are two sequences A and B with N (1<=N<=10000) elements each. All of the elements are positive integers. Given C=A*B, where '*' representing Cartesian product, c = a*b, where c belonging to C, a belonging to A and b belonging to B. Your job is to find the K'th largest element in C, where K begins with 1.


Input
Input file contains multiple test cases. The first line is the number of test cases. There are three lines in each test case. The first line of each case contains two integers N and K, then the second line represents elements in A and the following line represents elements in B. All integers are positive and no more than 10000. K may be more than 10000.


Output
For each case output the K'th largest number.


Sample Input
2
2 1
3 4
5 6
2 3
2 1
4 8
Sample Output
24
8


使用兩次二分

第一次十分枚舉第K大數，第二次二分判該第K大數合法性，根據是否有K-1個數比它小




源代碼：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

bool cmp(int a,int b)
{
        return a>b;
}

int find(int mid,int a[],int b[],int n)
{
        int res=0;
        int l=0;
        int r=n-1;
        for(int i=0;i<n;i++)
        {
                l=0;
                r=n-1;
                while(l<=r)
                {
                        int m=(l+r)/2;
                        if(a[i]*b[m]>=mid)
                        {
                                l=m+1;
                        }
                        else
                        {
                                r=m-1;
                        }
                }
                res+=l;
        }
        return res;
}

int main()
{
        int a[10010];
        int b[10010];
        int t;
        scanf("%d",&t);
        while(t--)
        {
                int n,k;
                scanf("%d%d",&n,&k);
                for(int i=0;i<n;i++)
                {
                        scanf("%d",&a[i]);
                }
                for(int i=0;i<n;i++)
                {
                        scanf("%d",&b[i]);
                }
                sort(a,a+n,cmp);
                sort(b,b+n,cmp);
                int left=a[0]*b[0];
                int right=a[n-1]*b[n-1];
                int m;
                int res;
                while(left>=right)
                {
                        int middle=(left+right)/2;
                        m=find(middle,a,b,n);
                        //printf("middle m: %d %d\n",middle,m);
                        if(m<k)
                        {
                                left=middle-1;
                        }
                        else
                        {
                                if(find(middle+1,a,b,n)<k)
                                {
                                        res=middle;
                                        break;
                                }
                                else
                                {
                                        right=middle+1;
                                }
                        }
                } printf("%d\n",res);
        }
        return 0;
}