51nod1105(二分)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1105

题意：中文题诶～

思路：直接二分答案，再通过二分找有多少组合大于等于当前值，确定答案；

代码：

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <math.h>
 #define ll long long
 using namespace std;

 const int MAXN=1e5;
 ll a[MAXN], b[MAXN];

 ll is_ok(ll mid, ll n){
     ll ans=;
     for(int i=; i<n; i++){
         // ll cnt=ceil(mid*1.0/a[i]);//***这里用ceil会wa，坑死，可能是double转long long有误差
         ll cnt=mid/a[i];
         if(mid%a[i]) cnt+=;
         int pos=lower_bound(b, b+n, cnt)-b;
         ans+=n-pos;
     }
     return ans;
 }

 int main(void){
     ll n, k;
     scanf("%lld%lld", &n, &k);
     for(int i=; i<n; i++){
         scanf("%lld%lld", &a[i], &b[i]);
     }
     sort(a, a+n);
     sort(b, b+n);
     ll l=a[]*b[], r=a[n-]*b[n-];
     while(l<=r){
         ll mid=(l+r)>>;
         ll cnt=is_ok(mid, n);
         if(cnt>=k) l=mid+;
         else r=mid-;
     }
     cout << l- << endl;
     return ;
 }