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

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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.

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

题目意思是，从一个n*m的乘法表（不要问我乘法表是什么）中选出第k小数（相同的数字会计算多次）。

比如样例 2 3 4

乘法表为

1 2 3

2 3 4

非减序列是：1, 2, 2, 3, 3, 4。第4个数字是3，所以输出3。

代码如下：

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL n, m, k;
LL min(LL a, LL b)
{
	return a<b?a:b;
}
LL check(LL x)//查找比x小的个数
{
	LL num = 0;
	for(int i = 1; i <= n; i++)
	{
		num+=min(m,x/i);
	}
	if(num >= k)
		return 1;
	else
		return 0;
}
int main()
{
	while(cin>>n>>m>>k)
	{
		LL l = 0, r = n*m, ans = 0;
		while(l <= r)
		{
			LL mid = (l+r)>>1;
			if(check(mid))
			{
				ans = mid;
				r = mid-1;
			}
			else
				l = mid+1;
		}
		cout<<ans<<endl;
	}
	return 0;
}