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

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 32 4 6

题意：一个n*m的矩阵。行乘上列等于这个点的值。问第k个大的数是多少。

题解：从l=1到r=n*m开始二分搜索mid=(l+r)/2;然后枚举，每一行有多少个小于mid的数，如果总数小于k则l=mid+1,否则r=mid;直到l>r终止。

#include <bits/stdc++.h> using namespace std;#define ll __int64ll n,m,k;ll cnt(ll s){ll t=0;for (int i=1;i<=n;i++){ll h=s/i;if (h>m)h=m;t+=h;}return t;}int main(){scanf("%I64d%I64d%I64d",&n,&m,&k);ll l=1;ll r=n*m+1;while (l<r){ll mid=(l+r)/2;if (cnt(mid)<k)l=mid+1;else r=mid;}cout<<r<<endl;}