二分法判断答案
1 second
256 megabytes
standard input
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.
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Print the k-th largest number in a n × m multiplication
table.
2 2 2
2
2 3 4
3
1 10 5
5
A 2 × 3 multiplication table looks like this:
1 2 3
2 4 6
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map> using namespace std; typedef long long int LL; LL n,m,k; bool ck(LL x)
{
LL nt=0;
for(LL i=1;i<=n;i++)
{
nt+=min(m,x/i);
}
if(nt>=k) return true;
return false;
} int main()
{
scanf("%I64d%I64d%I64d",&n,&m,&k); LL low=1,high=n*m,ans=-1,mid;
while(low<=high)
{
mid=(low+high)/2;
if(ck(mid))
{
ans=mid; high=mid-1;
}
else low=mid+1;
}
printf("%I64d\n",ans); return 0;
}
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