CF448D：Multiplication Table（二分）

最新推荐文章于 2020-08-27 17:46:34 发布

原创最新推荐文章于 2020-08-27 17:46:34 发布 · 381 阅读

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本文介绍了一种通过二分查找解决寻找n*m乘法表中第K大的数的问题的方法。利用二分查找技巧，遍历每行找到小于等于中间值的元素数量，并与K进行比较，最终确定目标值。

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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.

Examples

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：大概意思是一个n*m的二位数组，dp[i][j] = i*j，将数组所有的数升序排序后，求第k个数。

思路：从1~n*m二分，遍历每一行求有多少个数小于等于mid的跟k判断一下就行。

# include <stdio.h>
# include <algorithm>
using namespace std;
typedef long long LL;
LL n, m, k;
bool fun(LL num)
{
    LL ans = 0;
    for(LL i=1; i<=n; ++i)
    {
        LL imin = min(i*m, num);
        ans += imin/i;
    }
    return ans < k;
}
int main()
{
    while(~scanf("%I64d%I64d%I64d",&n,&m,&k))
    {
        LL l = 0, r = n*m, mid;
        while(l<r)
        {
            mid = (l+r)>>1;
            if(fun(mid))
                l = mid+1;
            else
                r = mid;
        }
        printf("%I64d\n",r);
    }
    return 0;
}