codeforces448D Multiplication Table(二分答案)

bigbigship發表於2014-07-20
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 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 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
我們要知道一個規律 ：在乘法表中第i行小於x的數有x/i個
因此在[1,m*n]內二分答案 即可
程式碼如下：
#include <iostream>

using namespace std;

typedef long long LL;

LL n,m,k;
bool judge(LL x)
{
    LL res=0;
    for(int i=1;i<=n;i++){
        LL tmp=min(i*m,x);
        res+=tmp/i;//第i行小於x數的個數為x/i;
    }
    return res<k;
}
LL binary_search(LL l,LL r)
{
    LL mid;
    while(l<r){
        mid=(l+r)>>1;
        if(judge(mid))
            l=mid+1;
        else
            r=mid;
    }
    return r;
}
int main()
{
    while(cin>>n>>m>>k){
        LL l=1,r=n*m;
        LL ans=binary_search(l,r);
        cout<<ans<<endl;
    }
    return 0;
}











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