HDU 4649 多校第五场1007

来源：互联网发布：漳州卓知职业培训学校编辑：程序博客网时间：2024/05/16 07:53

虽然这次做了5题，但是居然还是有板刷题没过掉。。

又是这节奏。。

贴一下这道板刷题。。

思路：每个数取他的二进制位，对于每一位，我们求他最后出现1的概率，那么最后的期望就是为1的概率乘以该位的十进制数，累加即可。

想到状压就是水题。。。

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <vector>#include <stack>#include <map>#include <iomanip>#define PI acos(-1.0)#define Max 2505#define inf 0x7fffffff#define LL(x) ( x << 1 )#define RR(x) ( x << 1 | 1 )#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define PII pair<int,int>#define bug puts("here")using namespace std;#define N 222double dp[22][N][2] ;//第i位第j个数是k的概率。int a[N] ;char b[N] ;double p[N] ;int n ;int main() {    int ca = 0 ;    while(cin >> n){        for (int i = 0 ; i <= n ; i ++ ){            scanf("%d",&a[i]) ;        }        for (int i = 1 ; i <= n ; i ++ ){            cin >> b[i] ;        }        for (int i = 1 ; i <= n ; i ++ ){            cin >> p[i] ;        }        for (int i = 0 ; i <= n ; i ++ ){            for (int j = 0 ; j <= 20 ; j++ ){                for (int k = 0 ;k < 2 ; k ++ )                    dp[j][i][k] = 0 ;            }        }        double ans = 0 ;        for (int i = 0 ; i <= 20 ; i ++ ){            if(a[0] & (1 << i))                dp[i][0][1] = 1 ;            else dp[i][0][0] = 1 ;            for (int j = 1 ; j <= n ; j ++ ){                dp[i][j][1] = dp[i][j - 1][1] * p[j] ;//该位不取                dp[i][j][0] = dp[i][j - 1][0] * p[j] ;//该位不取                if(a[j] & (1 << i)){//取该位                    if(b[j] == '&'){                        dp[i][j][1] += dp[i][j - 1][1] * (1 - p[j]) ;                        dp[i][j][0] += dp[i][j - 1][0] * (1 - p[j]) ;                    }                    else if(b[j] == '|'){                        dp[i][j][1] += dp[i][j - 1][1] * (1 - p[j]) ;                        dp[i][j][1] += dp[i][j - 1][0] * (1 - p[j]) ;                    }                    else {                        dp[i][j][1] += dp[i][j - 1][0] * (1 - p[j]) ;                        dp[i][j][0] += dp[i][j - 1][1] * (1 - p[j]) ;                    }                }                else {//同理                    if(b[j] == '&'){                        dp[i][j][0] += dp[i][j - 1][1] * (1 - p[j]) ;                        dp[i][j][0] += dp[i][j - 1][0] * (1 - p[j]) ;                    }                    else if(b[j] == '|'){                        dp[i][j][1] += dp[i][j - 1][1] * (1 - p[j]) ;                        dp[i][j][0] += dp[i][j - 1][0] * (1 - p[j]) ;                    }                    else {                        dp[i][j][0] += dp[i][j - 1][0] * (1 - p[j]) ;                        dp[i][j][1] += dp[i][j - 1][1] * (1 - p[j]) ;                    }                }            }            ans += (1 << i) * dp[i][n][1] ;        }        printf("Case %d:\n%.6f\n",++ca ,ans) ;    }    return 0 ;}