TYVJ-P1864 守卫者的挑战 题解

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<cstdlib>
 8 #include<iomanip>
 9 #include<cassert>
10 #include<climits>
11 #define maxn 10001
12 #define F(i,j,k) for(int i=j;i<=k;i++)
13 #define M(a,b) memset(a,b,sizeof(a))
14 #define FF(i,j,k) for(int i=j;i>=k;i--)
15 #define inf 0x7fffffff
16 #define q 23333333333333333
17 using namespace std;
18 int read(){
19     int x=0,f=1;char ch=getchar();
20     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
21     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
22     return x*f;
23 }
24 int a[210],n,m,t,i,j,k;
25 double p[210],f[201][201][401],ans;
26 int main()
27 {
28     std::ios::sync_with_stdio(false);
29 //  freopen("data.in","r",stdin);
30 //  freopen("data.out","w",stdout);
31     cin>>n>>m>>t;
32     F(i,1,n) cin>>p[i],p[i]/=100;
33     F(i,1,n) cin>>a[i];
34     if(t>n) t=n;
35     f[0][0][t+200]=1;
36     for(i=0;i<n;i++)
37      for(j=0;j<=n;j++)
38       for(k=-n;k<=n;k++)
39         if(a[i+1]==-1){
40             f[i+1][j][k+200]+=f[i][j][k+200]*(1-p[i+1]);
41             f[i+1][j+1][k-1+200]+=f[i][j][k+200]*p[i+1];
42         }
43         else{
44             f[i+1][j+1][min(k+a[i+1],n)+200]+=f[i][j][k+200]*p[i+1];
45             f[i+1][j][k+200]+=f[i][j][k+200]*(1-p[i+1]);
46         }
47     for(j=m;j<=n;j++)
48      for(k=0;k<=n;k++)
49         ans+=f[n][j][k+200];
50     cout<<setiosflags(ios::fixed)<<setprecision(6)<<ans<<endl;
51     return 0;        
52 }