CF 414B Mashmokh and ACM
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Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
3 2
5
6 4
39
2 1
2
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
动态规划思路:dp[i][j],I表示有I位的时候,J表示最后一个数字是J.( 初始化dp[0][1] = 1 ,是应为,在循环里面能把只有一位的时候都变成1,【1】,【2】,【3】,【4】。。。。。。也可以选择手动改吧I等于1的时候全都赋值成1)
#include <cstdio>#include <algorithm>#include <memory.h>#include <cmath>#include <cstring>using namespace std;#define P 1000000007int n, k, i, j, p, ans, dp[2100][2100];int main() {scanf("%d%d", &n, &k);dp[0][1] = 1;for (i = 1; i <= k; i++) {for (j = 1; j <= n; j++)for (p = j; p <= n; p += j)dp[i][p] = (dp[i][p] + dp[i - 1][j]) % P;}for (i = 1; i <= n; i++)ans = (ans + dp[k][i]) % P;printf("%d\n", ans);}
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