CF 240 D. Mashmokh and ACM

http://codeforces.com/contest/415/problem/D

题意：给出n和k，求长度为k的序列个数，序列满足1 <= b1 <= b2 ......<= bk <= n且任意一个数能整除它前面的数。

思路：dp[i][j]表示长度为i最后一位是j的序列个数。那么dp[i][j] =dp[i][j] + dp[i-1][k],其中k能被j整除.

#include <stdio.h>
#include <string.h>
#define LL __int64

const int mod = 1000000007;

LL dp[2010][2010];

int main()
{
	int n,k;
	while(~scanf("%d %d",&n,&k))
	{
		memset(dp,0,sizeof(dp));

		for(int i = 1; i <= n; i++)
			dp[1][i] = 1;

		for(int len = 1; len < k; len++)
		{
			for(int i = 1; i <= n; i++)
			{
				for(int j = 1; i*j <= n; j++)
				{
					dp[len+1][i*j] += dp[len][i];
					dp[len+1][i*j] %= mod;
				}
			}
		}

		LL res = 0;
		for(int i = 1; i <= n; i++)
		{
			res += dp[k][i];
			res %= mod;
		}
		printf("%I64d\n",res);
	}
	return 0;
}