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Codeforces Round #256 (Div. 2) D. Multiplication Table

2024-07-28 18:41:20   226人阅读

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
题意：让你在一个n*m的乘法矩阵中，找到第k大的数。

思路：二分查找，统计每一行的结果

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;

ll n, m, k;

int check(ll x) {
	ll res = 0;
	for (int i = 1; i <= n; i++) {
		ll tmp = min(i*m, x);
		res += tmp / i;
	}
	return res < k;
}

ll search(ll l, ll r) {
	while (l < r) {
		ll mid = (l + r) / 2;
		if (check(mid))
			l = mid + 1;
		else r = mid;
	}
	return r;
}

int main() {
	cin >> n >> m >> k;
	ll Right = n * m, Left = 1;
	ll ans = search(Left, Right);
	cout << ans << endl;
	return 0;
}


Codeforces Round #256 (Div. 2) D. Multiplication Table


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