Codeforces 448D： Multiplication Table（二分+数学）

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 3

2 4 6
题目大意：给你一个n*m的乘法表，问这个乘法表中第k大的数是几。
解题思路：二分，判断mid在表中第几大，然后缩短左右区间。怎么判断mid在表中第几大？
看这个例子：1 2 3 4 5 6
2 4 6 8 10 123 6 9 12 15 18 比如看10在这个表中第几大，第一行取6（min（列数，10/行数）），第二行取4（min（列数，10/行数）），第三行取3（min（列数，10/行数））。
代码如下：
#include <cstdio>
#include <algorithm>
using namespace std;
#define LL __int64
LL n,m,k;
bool judge(LL x)
{
LL sum=0;
for(LL i=1;i<=n;i++)//判断x在表中第几大
{
sum=sum+min(m,x/i);
}
if(sum>=k)
return true;
else
return false;
}
int main()
{
scanf("%I64d %I64d %I64d",&n,&m,&k);
LL l=1,r=n*m;
LL mid;
LL ans;
while(l<=r)
{
mid=(l+r)/2;
if(judge(mid))
{
ans=mid;
r=mid-1;
}
else
{
l=mid+1;
}
}
printf("%I64d\n",ans);
return 0;
}