Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) D. Jon and Orbs 概率DP

D. Jon and Orbs

链接：

http://codeforces.com/contest/768/problem/D

题解：

设dp[i][j]为第ｉ天时已经产生ｊ种龙晶，转移方程为

dp[i][j] += dp[i - 1][j] * j / k;
dp[i][j + 1] += dp[i - 1][j] * (k - j) / k;

代码：

 1 #include <map>
 2 #include <set>
 3 #include <cmath>
 4 #include <queue>
 5 #include <stack>
 6 #include <cstdio>
 7 #include <string>
 8 #include <vector>
 9 #include <cstring>
10 #include <iostream>
11 #include <algorithm>
12 #include <functional>
13 using namespace std;
14 #define rep(i,a,n) for (int i=a;i<=n;i++)
15 #define per(i,a,n) for (int i=n;i>=a;i--)
16 #define pb push_back
17 #define mp make_pair
18 #define all(x) (x).begin(),(x).end()
19 #define fi first
20 #define se second
21 #define SZ(x) ((int)(x).size())
22 typedef vector<int> VI;
23 typedef long long ll;
24 typedef pair<int, int> PII;
25 const ll mod = 1000000007;
26 const double eps = 1e-7;
27 // head
28 
29 double dp[10005][1005];
30 int main()
31 {
32     int k, q;
33     scanf("%d%d", &k, &q);
34     dp[0][0] = 1.0;
35     rep(i, 1, 10000) rep(j, 0, k) { 
36             dp[i][j] += dp[i - 1][j] * j / k;
37             dp[i][j + 1] += dp[i - 1][j] * (k - j) / k;
38         }
39     rep(i, 1, q) {
40         int p;
41         scanf("%d", &p);
42         rep(j, 0, 10000)
43             if (dp[j][k] > (p - eps) / 2000){
44                 printf("%d\n", j);
45                 break;
46             }
47     }
48     return 0;
49 }

 

posted @ 2017-02-22 18:54  Flowersea  阅读(215)  评论(0)  编辑  收藏  举报