Time Limit: 10000MS | Memory Limit: 64000K | |
Total Submissions: 2553 | Accepted: 1232 | |
Case Time Limit: 2000MS | Special Judge |
Description
Two companies, Macrosoft and Microhard are in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove that the program in question is disgusting. However, Ivan has a complicated problem. This new program has s subcomponents, and finding bugs of all types in each subcomponent would take too long before the target could be reached. So Ivan and Microhard agreed to use a simpler criteria --- Ivan should find at least one bug in each subsystem and at least one bug of each category.
Macrosoft knows about these plans and it wants to estimate the time that is required for Ivan to call its program disgusting. It's important because the company releases a new version soon, so it can correct its plans and release it quicker. Nobody would be interested in Ivan's opinion about the reliability of the obsolete version.
A bug found in the program can be of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bug cannot belong to two different categories or happen simultaneously in two different subsystems. The number of bugs in the program is almost infinite, so the probability of finding a new bug of some category in some subsystem does not reduce after finding any number of bugs of that category in that subsystem.
Find an average time (in days of Ivan's work) required to name the program disgusting.
Input
Output
Sample Input
1 2
Sample Output
3.0000
Source
題意:一共有n類bug,s段 subcomponents,某人每天能夠發現一個 subcomponent中的一個bug,求發現所有類bug並且每個 subcomponents中都有bug的期望。
思路:dp[i][j]表示找到i種bug,存在於j個subcomponents中的期望,則會發現,每當一天發現一個bug時,會有四種情況,將這些情況組合組合一下推出個公式帶進去就好了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
using namespace std;
const int mod=1e9+7;
const int N=1005;
double dp[N][N];
int main(){
int n,s;
scanf("%d%d",&n,&s);
dp[n][s]=0;
for(int i=n;i>=0;i--){
for(int j=s;j>=0;j--){
if(i==n&&j==s)continue;
dp[i][j]=(i*(s-j)*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(n-i)*(s-j)*dp[i+1][j+1]+n*s)/(n*s-i*j);
}
}
printf("%.4f\n",dp[0][0]);
return 0;
}