二分里面套二分

题目： https://www.51nod.com/Challenge/Problem.html#!#problemId=1105

思路：先二分答案，然后枚举一个数组，再二分另一个数组，非常典型的二分里面套二分

#include<bits/stdc++.h>
using namespace std;
ll a[500003];
ll b[500003];
int n,m;
int che(ll c)
{
    int sum=0;
    for(int i=n; i; i--)
    {
        ll z=c/a[i];
        if(c%a[i])z++;
        sum+=n-(lower_bound(b+1,b+1+n,z)-b-1);
    }
    return sum>=m;
}
int main()
{
    //ios::sync_with_stdio(false);

    cin>>n>>m;

    for(int i=1; i<=n; i++)
        sll(a[i],b[i]);
    sort(a+1,a+1+n);
    sort(b+1,b+1+n);
    ll l,r;
    l=a[1]*b[1];
    r=a[n]*b[n];
    ll qq=0;
    while(l<=r)
    {
        ll mid=(l+r)>>1;
        if(che(mid))qq=mid,l=mid+1;
        else r=mid-1;
    }
    cout<<qq<<endl;
}