CodeForces 448D Multiplication Table （二分）

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本文介绍了一个算法问题，即在一个N*M的乘法表中如何快速找到第K大的数。通过分析每行中小于特定数值X的元素数量，采用二分查找的方法高效解决问题。

K - Multiplication Table

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

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 Input

Input

2 2 2

Output

Input

2 3 4

Output

Input

1 10 5

Output

Hint

A 2 × 3 multiplication table looks like this:

1 2 3
2 4 6

题意：N*M的乘法表，找到第K个数

思路：第i行小于X的个数为min(（x-1）/i,m)

N行累加后与K进行比较

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
__int64 init(__int64 n,__int64 m,__int64 x)
{
	__int64 sum=0;
	for(int i=1;i<=n;i++)
	sum+=min((x-1)/i,m);
	return sum;
}
int main()
{
	__int64 n,m,k;
	while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
	{
		__int64 l=1;
		__int64 r=n*m;
		__int64 ans;
		while(l<=r)
		{
			__int64 mid=(l+r)/2;
			if(init(n,m,mid)<k)
			{
				l=mid+1;
				ans=mid;
			}
			else
			{
				r=mid-1;
			}
		}
		printf("%I64d\n",ans);
	}
	return 0;
}