Acwing 232. 守卫者的挑战 【概率】

#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <iostream>
#include <map>

#define go(i, l, r) for(int i = (l), i##end = (int)(r); i <= i##end; ++i)
#define god(i, r, l) for(int i = (r), i##end = (int)(l); i >= i##end; --i)
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define debug_in  freopen("in.txt","r",stdin)
#define debug_out freopen("out.txt","w",stdout);
#define pb push_back
#define all(x) x.begin(),x.end()
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pii;
const ll maxn = 1e5+10;
const ll maxM = 1e6+10;
const ll inf_int = 1e8;
const ll inf_ll = 1e17;

template<class T>void read(T &x){
    T s=0,w=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
    x = s*w;
}
template<class H, class... T> void read(H& h, T&... t) {
    read(h);
    read(t...);
}

void pt(){ cout<<'\n';}
template<class H, class ... T> void pt(H h,T... t){ cout<<" "<<h; pt(t...);}

//--------------------------------------------

double eps = 1e-7;
int N,L,K;
double E[205][205][405];
double p[205];
int a[205];
double dfs(int i,int j,int ca){ //在进行第i轮之前，成功轮j次，背包容量与碎片数量之差为ca
    if(i>N){//比赛结束后，看是否可以成功离开
        if(j>=L && ca>=0) return 1;
        else return 0;
    }
    if(E[i][j][ca + 200] >= 0) return E[i][j][ca + 200];
    double ans = 0;
    if(a[i] == -1) ans = p[i] * dfs(i+1,j+1,ca-1) + (1-p[i]) * dfs(i+1,j,ca); //情况1
    else ans = p[i] * dfs(i+1,j+1,ca+a[i]) + (1-p[i]) * dfs(i+1,j,ca); //情况2 . 其实可以合并一块写
    return E[i][j][ca + 200] = ans;
}
int main() {
//    debug_in;
//    debug_out;

    read(N,L,K); K = min(200,K);
    go(i,1,N) scanf("%lf",&p[i]),p[i]/=100;
    go(i,1,N) read(a[i]);

    go(i,0,200) go(j,0,200) go(k,0,400) E[i][j][k] = -1;
    double ans = dfs(1,0,K);
    printf("%.6f\n",ans);
    return 0;
}