题解【Codeforces414B】Mashmokh and ACM

题面

考虑设 \(dp_{i,j}\) 表示已经确定了前 \(i\) 个数，且最后一个数是 \(j\) 的方案数。

转移很容易想到：

• 设 \(v_j\) 表示 \(j\) 的所有约数的集合，那么就有 \(dp_{i,j}=\sum\limits_{k\in v_j}dp_{i-1,k}\)。

预处理出 \(v_j\)，然后直接转移即可。

#include <bits/stdc++.h>

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return f * x;
}

const int INF = 0x3f3f3f3f, N = 2003, M = N << 1, mod = 1000000007;

int n, m;
int dp[N][N];
vector <int> v[N];

int main()
{
    n = gi(), m = gi();
    for (int i = 1; i <= n; i+=1)
    	for (int j = 1; j * j <= i; j+=1)
    		if (i % j == 0)
    		{
    			v[i].push_back(j);
    			if (j * j != i) v[i].push_back(i / j);
    		}
    dp[0][1] = 1;
    for (int i = 1; i <= m; i+=1)
    	for (int j = 1; j <= n; j+=1)
    	{
    		for (auto x : v[j])
    			(dp[i][j] += dp[i - 1][x]) %= mod;
    	}
    int ans = 0;
    for (int i = 1; i <= n; i+=1)
    	(ans += dp[m][i]) %= mod;
    printf("%d\n", ans);
    return 0;
}