Background
If thou doest well, shalt thou not be accepted? and if thou doest not well, sin lieth at the door. And unto thee shall be his desire, and thou shalt rule over him.
And Cain talked with Abel his brother: and it came to pass, when they were in the field, that Cain rose up against Abel his brother, and slew him.
And the LORD said unto Cain, Where is Abel thy brother? And he said, I know not: Am I my brother's keeper?
And he said, What hast thou done? the voice of thy brother's blood crieth unto me from the ground.
And now art thou cursed from the earth, which hath opened her mouth to receive thy brother's blood from thy hand;
When thou tillest the ground, it shall not henceforth yield unto thee her strength; a fugitive and a vagabond shalt thou be in the earth.
—— Bible Chapter 4
Now Cain is unexpectedly trapped in a cave with N paths. Due to LORD's punishment, all the paths are zigzag and dangerous. The difficulty of the ith path is ci.
Then we define f as the fighting capacity of Cain. Every day, Cain will be sent to one of the N paths randomly.
Suppose Cain is in front of the ith path. He can successfully take ti days to escape from the cave as long as his fighting capacity f is larger than ci. Otherwise, he has to keep trying day after day. However, if Cain failed to escape, his fighting capacity would increase ci as the result of actual combat. (A kindly reminder: Cain will never died.)
As for ti, we can easily draw a conclusion that ti is closely related to ci. Let's use the following function to describe their relationship:
After D days, Cain finally escapes from the cave. Please output the expectation of D.
Input
The input consists of several cases. In each case, two positive integers N and f (n ≤ 100, f ≤ 10000) are given in the first line. The second line includes N positive integers ci (ci ≤ 10000, 1 ≤ i ≤ N)
Output
For each case, you should output the expectation(3 digits after the decimal point).
Sample Input
3 1 1 2 3
Sample Output
6.889
题目大意：有N条路，每条路都有一个Ci值，如今一个人要从中挑一条路逃出去。挑中每条路概率同样，一開始这个人有一个f值，他随机挑一条路。假设f值大于Ci那么他须要花费Ti天逃出去，假设小于的话那么就f增长Ci的大小，然后反复该过程，问逃出去须要的天数的数学期望
做法：DP[f] 表示 眼下这个人战斗力为f时逃出去须要的天数,转移就是n个嘛。。。
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <string> #include <algorithm> #include <queue> #include <cmath> using namespace std; const int maxn = 100000+10; struct path{ int c,t; path(int c,int t):c(c),t(t){} }; vector<path> vp; int n,f; double dp[maxn]; bool vis[maxn]; double dfs(int f){ if(vis[f]) return dp[f]; vis[f] = 1; double ans = 0; for(int i = 0; i < n; i++){ if(vp[i].c >= f){ ans += (1+dfs(f+vp[i].c))/n; }else{ ans += double(vp[i].t)/n; } } return dp[f] = ans; } int main(){ while(~scanf("%d%d",&n,&f)){ vp.clear(); memset(vis,0,sizeof vis); for(int i = 0; i < n; i++){ int t; scanf("%d",&t); int k = int((1+sqrt(5.0))/2*t*t); vp.push_back(path(t,k)); } printf("%.3lf ",dfs(f)); } return 0; }