CF 256# D. Multiplication Table 二分

http://codeforces.com/contest/448/problem/D

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 thei-th row and j-th column equalsi·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is thek-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 thek-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

思路： 数值很大的时候就要考虑二分了！ 2^30 大概就2 乘10的9次 ，所以25*10的10 的数，二分的次数不过才30多次；

时间复杂度差不多就 30 * o(10的5)

#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<list>#include<map>#include<set>using namespace std;typedef long long LL;LL n,m,k;bool check(LL x){    LL r=min(x,n);    LL cnt=0;    for(int i=1;i<=r;i++){        cnt+=min(x/i,m);    }    if(cnt>=k) return 1;    else return 0;}void deal(){    LL st=1,ed=n*m;    while(st<ed){        LL mid=(st+ed)>>1;        if(check(mid)) ed=mid;        else st=mid+1;    }    cout << st << endl;}int main(){//freopen("in.in","r",stdin);while(cin>>n>>m>>k){        deal();}return 0;}