cf448D. Multiplication Table【二分】

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 ann × 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.

Examples

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

确实没想到这题是二分。

二分判断条件对于每一个mid求出每一行比它小的数的个数和k判断

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=10010;
long long n,m,k;
bool judge(long long mid){
long long ans=0;
for(long long i=1;i<=n;++i){
ans=ans+min(m,mid/i);
}
return ans>=k;
}
int main()
{
scanf("%lld%lld%lld",&n,&m,&k);
long long l=1,r=n*m,ans;
while(l<=r){
long long mid=(l+r)>>1;
if(judge(mid)){
ans=mid;r=mid-1;
}
else {
l=mid+1;
}
}
printf("%lld\n",ans);
return 0;
}