Total Submission(s): 376 Accepted Submission(s): 199
As a result , Timer passed.
Now, you, the bad guy, also angered the Professor Tian when September Ends. You have to faced the problem too. The problem comes that there is an expression and you should calculate the excepted value of it. And the operators it may contains are '&' (and),'|'(or) and '^'(xor) which are all bit operators. For example: 7&3=3, 5&2=0, 2|5=7, 4|10=14, 6^5=3, 3^4=7.
Professor Tian declares that each operator O i with its coming number A i+1 may disappear, and the probability that it happens is P i (0<i<=n).
A i will be less than 2 20, 0<=P i<=1.
贴一下解题报告的思路:
反状态压缩——把数据转换成20位的01来进行运算
因为只有20位,而且&,|,^都不会进位,那么一位一位地看,每一位不是0就是1,这样求出每一位是1的概率,再乘以该位的十进制数,累加,就得到了总体的期望。
对于每一位,状态转移方程如下:
f[i][j]表示该位取前i个数,运算得到j(0或1)的概率是多少。
f[i][1]=f[i-1][1]*p[i]+根据不同运算符和第i位的值运算得到1的概率。
f[i][0]同理。
初始状态:f[0][0~1]=0或1(根据第一个数的该位来设置)
每一位为1的期望 f[n][1]
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #define N 250 using namespace std; int num[N]; double p[N],dp[N][2]; char opr[N],s1[N]; int main() { //freopen("data.in","r",stdin); int n,tem=1; char c1,c2; while(scanf("%d",&n)!=EOF) { for(int i=0;i<=n;i++) { scanf("%d",&num[i]); } getchar(); gets(s1); int l = strlen(s1); for(int i=0,j=0;i<=l-1;i++) { if(s1[i]=='|'||s1[i]=='^'||s1[i]=='&') { opr[j] = s1[i]; j++; } } for(int i=0;i<=n-1;i++) { scanf("%lf",&p[i]); } double ans=0.0; for(int i=1;i<=20;i++) { int t=num[0]&1; dp[0][t] = 1; dp[0][(t+1)%2] = 0; num[0]=(num[0]>>1); for(int j=1;j<=n;j++) { int k = num[j]&1; num[j]=(num[j]>>1); char c = opr[j-1]; if(c=='|') { if(k==1) { dp[j][1] = dp[j-1][1]+dp[j-1][0]*(1-p[j-1]); dp[j][0] = dp[j-1][0]*p[j-1]; }else { dp[j][1] = dp[j-1][1]; dp[j][0] = dp[j-1][0]; } continue; }else if(c=='&') { if(k==1) { dp[j][1] = dp[j-1][1]; dp[j][0] = dp[j-1][0]; }else { dp[j][1] = dp[j-1][1]*p[j-1]; dp[j][0] = dp[j-1][1]*(1-p[j-1])+dp[j-1][0]; } continue; }else if(c=='^') { if(k==1) { dp[j][1] = dp[j-1][0]*(1-p[j-1])+dp[j-1][1]*p[j-1]; dp[j][0] = dp[j-1][1]*(1-p[j-1])+dp[j-1][0]*p[j-1]; }else { dp[j][1] = dp[j-1][1]; dp[j][0] = dp[j-1][0]; } continue; } } ans += (1<<(i-1))*dp[n][1]; } printf("Case %d:\n",tem++); printf("%.6lf\n",ans); } return 0; }