nyoj1101Oh, my God!错排公式-白红宇的个人博客

nyoj1101Oh, my God!错排公式

发布日期：2021-06-29 11:14:05 浏览次数：2 分类：技术文章

本文共 1471 字，大约阅读时间需要 4 分钟。

nyoj1101Oh, 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 probabilitySee the following example.样例输入230样例输出Case [1]: 50.00%.Case [2]: 33.33%.

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         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;}

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