Codeforces Round #256 (Div. 2) D. Multiplication Table 二分法



D. Multiplication Table

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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

题意：给一个n*m的乘法表找第k大的数。

思路：二分。对于每一行找比mid小的数的个数加起来假设等于k就找到啦。

AC代码：

#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;

ll solve(ll x,int n,int m){
    ll res=0;
    for(int i=1;i<=n;i++){
        res+=min((ll)m,x/i);
    }
    return res;
}

int main(){
    int n,m;
    ll k;
    scanf("%d%d",&n,&m);
    cin>>k;
    ll l=1,r=(ll)n*m+1;
    while(l<r){
        ll mid=(l+r)/2;
        if(solve(mid,n,m)<k)
            l=mid+1;
        else
            r=mid;
    }
    cout<<l<<endl;
    return 0;
}