Multiplication Table (二分)

Bizon the Champion isn't just charming, he also is very smart.

While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.

Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·10^5; 1 ≤ k ≤ n·m).

Output
Print the k-th largest number in a n × m multiplication table.

Example
Input
2 2 2
Output
2
Input
2 3 4
Output
3
Input
1 10 5
Output
5
Note
A 2 × 3 multiplication table looks like this:


1 2 3

2 4 6

首先看一下题目，恩，还算简洁...（是不是有想法了？两重循环扫过去？）再看看数据...我去  这肯定超时啊啊！

怎么办呢？这种问题先往二分上凑一凑，先想一想，不是找非递增数列第k个数么，我每次二分一个数，看比他小的有多少个，与给的k比较，然后这样一直找下去，最后就可以得到第k个数啊！是不是很简单啊O(∩_∩)O

然后问题就是统计比二分出来的值小的个数，一行一行的统计，每行i比mid值小的数的个数为min(列数m,mid/i)

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<stack>#include<set>#include<queue>#include<cmath>#include<cstdlib>using namespace std;#define INF 999999typedef long long ll;const int maxn=1500;int main(){    ll n,m,k;    while(scanf("%lld%lld%lld",&n,&m,&k)!=EOF){        ll l=1,r=m*n,mid;        while(l!=r){            mid=(l+r)/2;            ll sum=0;            for(int i=1;i<=n;i++){                ll x=mid/i;                if(x<=m)sum+=x;                else sum+=m;            }            if(sum<k)l=mid+1;            else r=mid;        }        printf("%lld\n",l);    }    return 0;}