nyoj1101Oh, my God!错排公式

nyoj1101
Oh, my God!
时间限制：1000 ms  |  内存限制：65535 KB
难度：2
描述
In order to happy everyone, organizer HRW held an open up 
partythere have specific requirements for this activity is 
this:
First of all, all person in the party will have to write a 
note to his name into the box;Then, after all the note added 
is completed, each taking a note from the box;Finally, if 
made note of your name, then "Congratulations, won the party!"
Oh, my God!
Now to the question, you can calculate the probability of 
this happening?Don't count? Don't you want said by others as 
a  DouBi?!  AC it！
输入
Input file contains several test cases. Each test case consists 
of a integer numbers n on a line(one≤n≤ten ).The last tes case
 is followed by a line that contains one zeroes. This line 
 must not be processed.
输出
print the the probability
See the following example.
样例输入
2
3
0
样例输出
Case [1]: 50.00%.
Case [2]: 33.33%.

#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<map>
using namespace std;
//错排公式 
//https://www.cnblogs.com/c1299401227/p/5349727.html 
int main() {
  int n;
  int count_ = 1;
  double dp[11], dp2[11];
  dp2[0] = dp2[1] = 1, dp2[2] = 2;
  dp[1] = 0, dp[2] = 1;
  for (int i = 3; i <= 10; i++) {
    dp2[i] = dp2[i - 1] * i;
    dp[i] = (i - 1) * (dp[i - 2] + dp[i - 1]);
  }
  while (scanf("%d", &n), n) {
    if (n != 1) printf("Case [%d]: %.2f%%.\n", count_, 100 * (dp[n] / dp2[n]));
    else printf("Case [%d]: 100.00%%.\n", count_);
    count_++;
  }
  return 0;
}