【组队赛三】-E Binary Search cf448D

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

Status

Practice

CodeForces 448D
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 3
2 4 6

<span style="color:#3333ff;"><span style="color:#3333ff;background-color: rgb(255, 255, 255);">/*
_______________________________________________________________________________________

       author    :   Grant yuan
       time      :   2014.7.21
       algorithm :   Binary Search
       explain   :   如果i*m<=aa,则会有m个数满足结果，否则会有aa/i个数满足结果 
________________________________________________________________________________________
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<functional>
#define INF 999999999
using namespace std;

//long long a[100003][100003];
long long k;
long long l,r,mid;
long long n,m;
long long M;

inline bool can(long long aa)
{   long long sum=0;
    for(int i=1;i<=n;i++)
         {
             if(i*m<=aa)
                sum+=m;
             else sum+=aa/i;
         }
   if(sum>=k)
       return true;
   return false;
}

int main()
{   M=0;
    cin>>n>>m>>k;
        long long ans=1;
        l=1;r=n*m;
        while(l<=r){
            mid=(long long)((l+r)*0.5);
            if(can(mid))
            {  ans=mid;
                r=mid-1;
            }
            else
                l=mid+1;
        }
          cout<<ans<<endl;
}
</span></span>