HDU 2068 RPG的错排

 

http://acm.hdu.edu.cn/showproblem.php?pid=2068

 

解题思路：错排+组合C(n,n/2)，累加i=0-->i=n/2的（错排*组合C(n,i)）的值

 

#include <stdio.h>int main(void){ int i, n; double s; double d[26][2] = {{1,1},{1,0},{2,1},{6,2}}; for (i = 4; i < 26; i++) { d[i][0] = i * d[i-1][0]; d[i][1] = (i - 1) * (d[i-1][1] + d[i-2][1]); } while (scanf("%d", &n), n) { for (s = 0, i = n/2; i >= 0; i--) s += 1.0*d[n][0]/d[i][0]/d[n-i][0]*d[i][1]; printf("%.lf/n", s); }}