UVA11246 K-Multiple Free set【数学+找规律】

A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another
integer multiplied by k. For example for k = 2, {1,3,4} is a valid set, but not {2,4,5}, as 4 is double of 2.
    You will be given n and k. you have to determine the largest k-multiple free subset of the integers from 1 to n.

Input
First line of the input contains T (1 ≤ T ≤ 1000) the number of test case. Then following lines contains T test cases. Each case contains a line containing 2 integers n (1 ≤ n ≤ 1000000000) and k (2 ≤ k ≤ 100).

Output
For each test case output contains 1 integer the size of the largest k-multiple free subset of the integers from 1 to n.

Sample Input
3
10 2
100 2
1000 2
Sample Output
6
67
666

问题链接：UVA11246 K-Multiple Free set
问题简述：一个{1…n}的集合，计算其子集合，使得元素个数最多，并且不存在有两个元素x1 * k = x2，求出最多的元素个数是多少？
问题分析：
    开始时子集有n个元素，先删除k倍的，删除为{k, 2k, 3k, 4k, 5k, 6k…}，会删掉多余的k²，因此需要再加回k²倍的数。这时候会出现多出k²的情况，如迭代下去，得到最后答案为n - n / k + n / (k ^ 2) - n / (k ^ 3) + n / (k ^ 4)…。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA11246 K-Multiple Free set */

#include <bits/stdc++.h>

using namespace std;

int solve(int n, int k)
{
    int sign = 1, ans = 0;
    while (n) {
        ans += sign * n;
        n /= k;
        sign = - sign;
    }
    return ans;
}

int main()
{
    int t, n, k;

    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &k);
        printf("%d\n", solve(n, k));
    }

    return 0;
}