Codeforces 448D Multiplication Table

life4711發表於2014-07-19

http://codeforces.com/problemset/problem/448/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 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
題目大意:給定一個二維陣列，a[i][j]=i*j;求出第k大的數；

解題思路：二分1~n*m，對於每一個mid求出沒一行比不比它大的數的個數ret，它就是第ret大的數，然而對於某些數可能不存在，我仔細想過了，如果一個列舉到不存在的數，那麼比它小的數中一定還有滿足的數而且存在於這個二維陣列中，因此我們二分結束的標誌是l<r，找到滿足的那麼就在1~ret中繼續尋找。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stdio.h>
using namespace std;
long long n,m,k;
int main()
{
   while(~scanf("%lld%lld%lld",&n,&m,&k))
   {
       long long  l=1,r=n*m;
       while(l<r)
       {
             long long  mid=(l+r)/2;
             long long  ret=0;
             for(long long i=1;i<=n;i++)
                ret+=min(m,mid/i);
             if(ret>=k)
                 r=mid;
             else
                 l=mid+1;
       }
       printf("%lld\n",r);
   }
   return 0;
}


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