Aimee

想出状态转移的难度很小

很强的题解

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=2020;
double eps=1e-5;
double Aimee[maxn];
double p[maxn];
double f[2][maxn];
int n,m,k;
double p1,p2,p3,p4;
int main(){
    while(scanf("%d%d%d%lf%lf%lf%lf",&n,&m,&k,&p1,&p2,&p3,&p4)!=EOF){
        if(p4<eps){
            printf("0.00000n");
            continue;
        }
        double p21=p2/(1-p1);
        double p41=p4/(1-p1);
        double p31=p3/(1-p1);
        p[0]=1.0;
        for(int i=1;i<=n;++i) p[i]=p21*p[i-1];
        f[1][1]=p41/(1-p21);
        Aimee[1]=p41;
        for(int i=2;i<=n;++i){
            int Ai=i&1;
            for(int j=2;j<=k;++j) 
                Aimee[j]=p31*f[Ai^1][j-1]+p41;
            for(int j=k+1;j<=i;++j){
                Aimee[j]=p31*f[Ai^1][j-1];
            }
            double tmp=Aimee[1]*p[i-1];
            for(int j=2;j<=k;j++)
                tmp+=Aimee[j]*p[i-j];
            for(int j=k+1;j<=i;++j)
                tmp+=Aimee[j]*p[i-j];
            f[Ai][i]=tmp/(1-p[i]);
            f[Ai][1]=p21*f[Ai][i]+Aimee[1];
            for(int j=2;j<i;++j) f[Ai][j]=p21*f[Ai][j-1]+Aimee[j];
        }
        printf("%.5lfn",f[n&1][m]);
    }
    return 0;
}