lightoj1038 - Race to 1 Again(概率DP)

来源：互联网发布：linux云计算就业前景编辑：程序博客网时间：2024/06/02 21:19

题意：一开始给你一个数D然后随机从它的因子中找出一个x，进行D = D / x操作，这样算一步，直到D=1结束，问你P(D)的期望是几步。

思路：建边之后就是典型的概率DP题，状态转移方程式： P(D) = (∑(P(D的非1和本身因子)+1) + 2) / (因子数+1) ；

#include <cstdio>#include <cstring>const int maxn = 100005;const double eps = 1e-7;struct Edge{      int to, next;}edge[maxn*20];double dp[maxn];int E, head[maxn];void add_edge(int u, int to){      edge[E].to = to;      edge[E].next = head[u];      head[u] = E++;}void init(){      E = 0;      memset(head, -1, sizeof(head));      for (int i = 0; i < maxn; i++)dp[i] = 0.0;      for (int x = 100000; x >= 2; x--){            for (int i = 2; i * i <= x; i++)if (x % i == 0){                  add_edge(x, i);                  if (x / i != i)add_edge(x, x/i);            }      }}double dfs(int u){      if (dp[u] >= eps)return dp[u];      if (u == 1)return 0.0;      double sum = 2;      int cnt = 2;      for (int i = head[u]; i != -1; i = edge[i].next){            cnt++;            sum += dfs(edge[i].to) + 1;      }      return dp[u] = sum / (cnt-1);}int main(){      int T, n;      init();      scanf("%d", &T);      for (int cas = 1; cas <= T; cas++){            scanf("%d", &n);            printf("Case %d: %f\n", cas, dfs(n));      }      return 0;}