Hdu-4649-worker sor Tian

--> The operand range is <2 ^ 20, which indicates bitwise operation. each operand can be converted into a binary number of up to 20 bits, after the number of n + 1 Operations is obtained, the probability of 1 appears for each binary bit (the probability of 0 does not need to be calculated. If this bit is 0, the total number is irrelevant to this bit, only occupies space ).

Set a [I] [j] to represent the j-th binary bit of the I-th number; p [I] [j] to represent the bit of the binary, after the first I count operation, the probability of j is obtained (j is 0 or 1). d [I] indicates the probability that the I th binary bit appears 1 after the n + 1 count operation.

For p [I] [j], the state transition equation is as follows:

P [I] [1] = p [I-1] [1] * P [I] + probabilities produced by various computations.

P [I] [0] = 1-p [I] [1].

 

#include <cstdio>using namespace std;const int maxn = 200 + 3;const int maxm = 20 + 3;int A[maxn], a[maxn][maxm], n;char O[maxn];double P[maxn], p[maxn][2], d[maxm];void read(){    int i;    for(i = 0; i < n+1; i++) scanf("%d", &A[i]);    for(i = 1; i <= n; i++){        getchar();        O[i] = getchar();    }    for(i = 1; i <= n; i++) scanf("%lf", &P[i]);}void init(){    int i, j;    for(i = 0; i < n+1; i++)        for(j = 0; j < 20; j++) a[i][j] = (1 << j) & A[i];}void dp(){    int i, bit;    for(bit = 0; bit < 20; bit++){        if(a[0][bit]){            p[0][1] = 1;            p[0][0] = 0;        }        else{            p[0][0] = 1;            p[0][1] = 0;        }        for(i = 1; i < n+1; i++){            p[i][1] = p[i-1][1] * P[i];            switch(O[i]){                case '&':{                    if(a[i][bit]) p[i][1] += p[i-1][1] * (1 - P[i]);                    break;                }                case '|':{                    if(a[i][bit]) p[i][1] += 1 - P[i];                    else p[i][1] += p[i-1][1] * (1 - P[i]);                    break;                }                case '^':{                    if(a[i][bit]) p[i][1] += p[i-1][0] * (1 - P[i]);                    else p[i][1] += p[i-1][1] * (1 - P[i]);                    break;                }            }            p[i][0] = 1 - p[i][1];        }        d[bit] = p[n][1];    }}double solve(){    double ret = 0;    for(int i = 0; i < 20; i++) ret += (1 << i) * d[i];    return ret;}int main(){    int kase = 1;    while(scanf("%d", &n) == 1){        read();        init();        dp();        printf("Case %d:\n%.6lf\n", kase++, solve());    }    return 0;}