Multiplication Table - #256 (Div. 2) D (448D) 二分

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本文介绍了一个算法，用于在给定大小的乘法表中找到第K大的数。通过二分查找，该算法能高效地解决这个问题。

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

题意：找到第几小的数。

思路：二分。直接暴力二分………………

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll n,m,k,l,r,mid;
ll solve(ll num)
{ ll i,j,sum=0,ret;
  for(i=1;i<=n;i++)
  { ret=num/i;
    if(ret>m)
     ret=m;
    sum+=ret;
  }
  return sum;
}
int main()
{ scanf("%I64d%I64d%I64d",&n,&m,&k);
  l=1;r=n*m;
  while(l<r)
  { mid=(l+r)/2;
    if(solve(mid)<k)
     l=mid+1;
    else
     r=mid;
  }
  printf("%I64d\n",r);
}