SSL 2386 序列#线性动态规划#

比赛

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题目

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分析

dp，状态转移方程：f[i][j]表示第j位的数字是i的方案$f[i][j]表示第j位的数字是i的方案$
f[i][j]+=f[(i的因数)][j−1]（逆推）$f[i][j]+=f[(i的因数)][j-1]（逆推）$
f[i∗(1ton/j)][j]+=f[i][j−1](顺推)$f[i\ast (1_{to}n/j)][j]+=f[i][j-1](顺推)$

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逆推代码

#include <cstdio>
#include <cstring>
using namespace std;
struct node{int x,y,next;}e[100001];
int f[2002][2002],n,m,k,ans,ls[2001];
int main(){
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++) f[i][1]=1;
    for (int i=1;i<=n;i++)
    for (int j=1;j<=i;j++)
    if (i%j==0) e[++m].x=i,e[m].y=j,e[m].next=ls[e[m].x],ls[e[m].x]=m;//邻接表优化
    for (int j=2;j<=k;j++)
    for (int i=1;i<=n;i++){
    int t=ls[i];
    while (t){
        f[i][j]=(f[i][j]+f[e[t].y][j-1])%1000000007;
        t=e[t].next;
    }
    }
    for (int i=1;i<=n;i++) ans=(ans+f[i][k])%1000000007;
    printf("%d",ans);
    return 0;
} 

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顺推代码

#include <cstdio>
#include <cstring>
#define p 1000000007
using namespace std;
int f[2002][2002],n,m,ans;
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++) f[i][1]=1;
    for (int j=2;j<=m;j++)
    for (int i=1;i<=n;i++)
    for (int k=1;k<=n/i;k++) f[i*k][j]=(f[i][j-1]+f[i*k][j])%p;
    for (int i=1;i<=n;i++) ans=(ans+f[i][m])%p;
    printf("%d",ans);
    return 0;
}