UVA&&POJ离散概率与数学期望入门练习[4]

POJ3869 Headshot

题意：给出左轮手枪的子弹序列，打了一枪没子弹，要使下一枪也没子弹概率最大应该rotate还是shoot

条件概率，|00|/(|00|+|01|)和|0|/n谁大的问题

|00|+|01|=|0|

注意序列是环形

//
//  main.cpp
//  poj3869
//
//  Created by Candy on 25/10/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=;
int n,a,b;
char s[N];
int main(int argc, const char * argv[]) {
    scanf("%s",s+);
    n=strlen(s+);
    for(int i=;i<=n;i++){
        if(s[i]=='') b++;
        if(s[i]==''&&s[i+]=='') a++;
    }
    if(s[n]==''&&s[]=='') a++;
    if(a*n>b*b) printf("SHOOT");
    else if(a*n==b*b) printf("EQUAL");
    else printf("ROTATE");
    return ;
}

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UVA - 10491

Cows and Cars

经典问题，a奶牛，b车，c门展示

全概率公式，分成一开始选了牛a/(a+b)和一开始选了车b/(a+b)两部分

部分里总是换门得到车概率分别是b/(a+b-c-1)和(b-1)/...

#include <cstdio>
double a,b,c;
int main(int argc, const char * argv[]) {
    while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF)
        printf("%.5f\n",(a*b+b*(b-))/((a+b)*(a+b-c-)));
    return ;
}

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UVA - 11181

Probability|Given

题意：n个人，买东西概率pi，有r个人买了东西，求每个人实际买东西概率

条件概率

E为r个人买东西，Ei为r个人中有i买东西

P(Ei|E)=P(EiE)/P(E)

计算概率用dfs爆搜每个人买还是不买即可

PS：不要读入优化，浮点数

//
//  main.cpp
//  uva11181
//
//  Created by Candy on 25/10/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=;
typedef long long ll;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}
int n,r,vis[N];
double p[N],pe[N];
void dfs(int d,int cnt,double prob){//printf("dfs %d %d %f\n",d,cnt,prob);
    if(cnt>r||d--cnt>n-r) return;
    if(d==n+){
        if(cnt==r)
            for(int i=;i<=n;i++) if(vis[i]) pe[i]+=prob;
        pe[]+=prob;
        return;
    }
    vis[d]=;
    dfs(d+,cnt+,prob*p[d]);
    vis[d]=;
    dfs(d+,cnt,prob*(-p[d]));
}
int main(int argc, const char * argv[]){
    int cas=;
    while((n=read())){printf("Case %d:\n",++cas);
        r=read();
        for(int i=;i<=n;i++) scanf("%lf",&p[i]);
        memset(vis,,sizeof(vis));
        memset(pe,,sizeof(pe));
        dfs(,,1.0);
        for(int i=;i<=n;i++) printf("%.6f\n",pe[i]/pe[]);
    }

    return ;
}

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UVA - 12230

Crossing Rivers

每条河的时间均匀分布在l/v~3*l/v 期望过河时间就是2*l/v

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int n;
double d,p,l,v,sum,ans;
int main(int argc, const char * argv[]){
    int cas=;
    while(scanf("%d%lf",&n,&d)!=EOF&&(n||d)){
        ans=sum=;
        for(int i=;i<=n;i++){
            scanf("%lf%lf%lf",&p,&l,&v);
            sum+=l;
            ans+=*l/v;
        }
        printf("Case %d: %.3f\n\n",++cas,ans+d-sum);
    }

    return ;
}