Multiplication Table（二分搜索）

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

 

        题意：

        给出 n，m（1 ~ 5 * 10 ^ 5 ），k （1 ~ n * m ），代表有 n 行 m 列的数据，矩阵中的值为 i * j。求出矩阵中第 k 个大的数。

 

        思路：

        二分搜索。取区间为左开右闭区间，扫描每一行 num / i 的值有几个，算出的这个总数如果大于 k，不能说明这个数不是第 k 个大的数，因为可能会有重复的，但是如果这个总数小于 k，一定能说明这个数并非第 K 个大的数，所以这时候应该往右边搜，最后搜出来右区间的值即为答案。还有细节地方要处理就是 扫描每一行的时候要进行 min（num / i ， m）操作，因为有可能除出来的数大于列数，那么加上的应该是这一行的列数个，而不是除数个。

 

        AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

ll n, m, k, l, r;

ll C (ll num) {
        ll ans = 0;

        for (ll i = 1; i <= n; ++i) 
                ans += min((num / i), m);

        return ans;
}

void bit_search () {

        while (r - l > 1) {
                ll mid = l + (r - l) / 2;

                if (C(mid) < k) l = mid;
                else r = mid;
        }

        printf("%I64d\n", r);
}

int main() {

        scanf("%I64d%I64d%I64d", &n, &m, &k);

        l = 0;
        r = n * m;

        bit_search();

        return 0;
}