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：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 #include <math.h>
 5 #define ll long long
 6 using namespace std;
 7 
 8 const int MAXN=1e5;
 9 ll a[MAXN], b[MAXN];
10 
11 ll is_ok(ll mid, ll n){
12     ll ans=0;
13     for(int i=0; i<n; i++){
14         // 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
15         ll cnt=mid/a[i];
16         if(mid%a[i]) cnt+=1;
17         int pos=lower_bound(b, b+n, cnt)-b;
18         ans+=n-pos;
19     }
20     return ans;
21 }
22 
23 int main(void){
24     ll n, k;
25     scanf("%lld%lld", &n, &k);
26     for(int i=0; i<n; i++){
27         scanf("%lld%lld", &a[i], &b[i]);
28     }
29     sort(a, a+n);
30     sort(b, b+n);
31     ll l=a[0]*b[0], r=a[n-1]*b[n-1];
32     while(l<=r){
33         ll mid=(l+r)>>1;
34         ll cnt=is_ok(mid, n);
35         if(cnt>=k) l=mid+1;
36         else r=mid-1;
37     }
38     cout << l-1 << endl;
39     return 0;
40 }

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