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社区首页 >专栏 >【CodeForces】448D - Multiplication Table(二分)

【CodeForces】448D - Multiplication Table(二分)

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FishWang
发布2025-08-26 20:32:01
发布2025-08-26 20:32:01
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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.

Examples

input

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2 2 2

output

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2

input

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2 3 4

output

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3

input

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1 10 5

output

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5

Note

A 2 × 3 multiplication table looks like this:

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1 2 3
2 4 6

从 1 到 n*m 二分找结果。注意数据类型。

代码如下:

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#include <cstdio>
#include <algorithm>
using namespace std;
__int64 n,m,k;
bool check(__int64 x)
{
	__int64 sum = 0;
	for (int i = 1 ; i <= n ; i++)
	{
		sum += min(m , x / i);
		if (sum >= k)
			return true;
	}
	return false;
}
int main()
{
	scanf ("%I64d %I64d %I64d",&n,&m,&k);
	__int64 l,r,mid;
	l = 1;
	r = n*m;
	while (r >= l)
	{
		mid = (l + r) / 2;
		if (check(mid))
			r = mid - 1;
		else
			l = mid + 1;
	}
	printf ("%I64d\n",l);
	return 0;
}
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原始发表:2025-08-26,如有侵权请联系 cloudcommunity@tencent.com 删除

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