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CodeForces 415D Mashmokh and ACM

$dp$。

记$dp[i][j]$表示已经放了$i$个数字,并且第$i$个数字放了$j$的方案数。那么$dp[i][j] = \sum\limits_{k|j}^{}  {dp[i - 1][k]}$

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<iostream>using namespace std;typedef long long LL;const double pi=acos(-1.0),eps=1e-6;void File(){    freopen("D:\\in.txt","r",stdin);    freopen("D:\\out.txt","w",stdout);}template <class T>inline void read(T &x){    char c = getchar();    x = 0;    while(!isdigit(c)) c = getchar();    while(isdigit(c)) { x = x * 10 + c - 0; c = getchar(); }}const int maxn=2020;LL dp[maxn][maxn],mod=1e9+7;int n,m;int main(){    scanf("%d%d",&m,&n);    for(int i=1;i<=m;i++) dp[1][i]=1;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            for(int k=j;k<=m;k=k+j)            {                dp[i+1][k]=(dp[i+1][k]+dp[i][j])%mod;            }        }    }    LL ans=0;    for(int i=1;i<=m;i++) ans=(ans+dp[n][i])%mod;    cout<<ans<<endl;    return 0;}

 

CodeForces 415D Mashmokh and ACM