Codeforces 448D. Multiplication Table

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 ≤ 5·105; 1 ≤ k ≤ n·m).

Output

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

二分答案，每次判断当前数多大的函数是O(n)，二分次数应为O(lognm)次，所以总共O(nlongnm)

ps: Python跑的好慢啊。。看来c++要跑150ms的Python就得超时

#include<bits/stdc++.h>  
using namespace std;
typedef long long LL;   


LL n,m,k,l,r;
LL f(LL now){
    LL res = 0;
    now = now - 1;
    for (int i=1;i<=n;i++)
        res += min(m,now/i);
    return res;
}
int main(){  
    // freopen("test.in","r",stdin);
    cin >> n >> m >> k;
    l = 1,r = n*m+1;
    while (l < r){
        LL mid = (l + r) / 2;
        if (f(mid) < k) l = mid+1;
        else r = mid;
    }
    cout << l-1;
    return 0;   
}