Codeforces 414B Mashmokh and ACM

http://codeforces.com/problemset/problem/414/B

题目大意：

题意:一个序列B1,B2...Bl如果是好的,必须满足Bi | Bi + 1(a | b 代表a整除b), 求长度为K,元素大小小于等于N的序列个数

思路:f[i][j] 代表在序列的第i位，当前数字为j的方案数

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
const int Mod=1000000007;
int f[5005][5005];
int n,K;
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
int main(){
    n=read();K=read();
    for (int i=1;i<=n;i++)
     f[1][i]=1;
    for (int i=2;i<=K;i++)
     for (int j=1;j<=n;j++){
        for (int k=1;k<=sqrt(j);k++)
         if (j%k==0){
                f[i][j]=(f[i][j]+f[i-1][k])%Mod;
                if (k*k!=j)
                f[i][j]=(f[i][j]+f[i-1][j/k])%Mod;
         }
     }
    int ans=0;
    for (int i=1;i<=n;i++)
     ans=(ans+f[K][i])%Mod;
    printf("%d
",ans);  
    return 0;
}