HDU 4465 递推与double的精确性

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 const int N = 200005;
 6 double dp[N][2];
 7 
 8 int main()
 9 {
10    // freopen("test.in","rb",stdin);
11 
12     int n,cas = 0;
13     double p;
14     while(scanf("%d%lf",&n,&p)!=EOF){
15         double ans1 = 0;
16         double ans2 = 0;
17         int last[2];
18         last[0] = last[1] = n+1;
19         double q = 1-p;
20         dp[n][0] = 1;
21         dp[n][1] = 1;
22         ans1 += n;
23         ans2 += n;
24         for(int i=n-1;i>=1;i--){
25             double tmp1 =  q * (2*n-i) / (n-i);
26             double tmp2 =  p * (2*n-i) / (n-i);
27             
28             //递推过程，因为数目太多，p直接开方，数据大点的话就直接因精确度不够变为0
29             //所以每次在答案超过总数的情况下乘个p值
30             dp[i][0] = tmp1 * dp[i+1][0];
31             dp[i][1] = tmp2 * dp[i+1][1];
32             ans1 += (dp[i][0] * i);
33             while(ans1>n){
34                 dp[i][0] *= p;
35                 ans1 *= p;
36                 last[0]--;
37             }
38             ans2 += (dp[i][1] * i);
39             while(ans2>n){
40                 dp[i][1]*=q;
41                 ans2 *= q;
42                 last[1]--;
43             }
44         }
45        // cout<<"  dp "<<dp[n]<<endl;
46         ans1 *= pow(p,last[0]);
47         ans2 *= pow(q,last[1]);
48         cas++;
49 
50         printf("Case %d: %.6f\n",cas,ans1+ans2);
51     }
52     return 0;
53 }