codeforce 448D Multiplication Table

 Multiplication Table

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

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 an n × 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

2

Input

2 3 4

Output

3

Input

1 10 5

Output

5

Hint

A 2 × 3 multiplication table looks like this:

1 2 32 4 6

二分1~n*m，计算每行小于等于mid的数，符合K个即可。

代码：

#include<cstdio>long long min(long long x,long long y){if(x<y)return x;return y;  } long long n,m,k;bool judge(long long mid){long long sum=0;for(long long i=1;i<=n;i++)//坐标要从1开始。。。0不能作除数………… {sum+=min(m,mid/i);}return sum>=k; } int main(){scanf("%lld%lld%lld",&n,&m,&k);long long left=1,right=n*m,mid;long long ans; while(left<=right){mid=(left+right)/2;if(judge(mid)){ans=mid;right=mid-1;}elseleft=mid+1;}printf("%lld\n",ans);return 0;}

查找算法现在就学了二分，暴力或者想不出来方法的时候，多想一下二分吧，或许有效