Geek's Collection(幂运算)

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3625

View Code
 1 #include <stdio.h>
 2 #include <math.h>
 3 #define E 0.57721566490153286060651209
 4 int main()
 5 {
 6     double a,x;
 7     while (scanf("%lf",&a) != EOF)
 8     {
 9 
10         x = E;
11         x *= pow(2,a) - 1;
12         printf("%1.12e\n",x);
13     }
14     return 0 ;
15 }
原文地址:https://www.cnblogs.com/yelan/p/2992250.html