题解 | #简单题2#

有了第一题的经验，我们可以立即猜出所谓 $\gamma =e=2.71828\cdots \backslash gamma=e=2.71828\backslash cdots$

而事实上这道题也就是把式子模拟一下就可以了：

$a^b=c\times da^b=c\backslash times\; d$

所以 $d={\displaystyle \frac{a^b}{c}}d=\backslash dfrac\{a^b\}\{c\}$（为了方便书写，我没用题目中的奇怪字母了）

#include<cmath>
#include<cstdio>
int init(){
	char c = getchar();
	int x = 0, f = 1;
	for (; c < '0' || c > '9'; c = getchar())
		if (c == '-') f = -1;
	for (; c >= '0' && c <= '9'; c = getchar())
		x = (x << 1) + (x << 3) + (c ^ 48);
	return x * f;
}
void print(int x){
	if (x < 0) x = -x, putchar('-');
	if (x > 9) print(x / 10);
	putchar(x % 10 + '0');
}
int main(){
	int T = init();
	while (T--) {
		double alpha = (double) init(), beta = (double) init();
		int gamma = init();
		double theta = 2.7182818284590452;
		switch (gamma) {
			case 1 : { printf("%.1lf\n", pow(alpha, theta) / beta); break; }
			case 2 : { printf("%.2lf\n", pow(alpha, theta) / beta); break; }
			case 3 : { printf("%.3lf\n", pow(alpha, theta) / beta); break; }
			case 4 : { printf("%.4lf\n", pow(alpha, theta) / beta); break; }
			case 5 : { printf("%.5lf\n", pow(alpha, theta) / beta); break; }
		}
	}
}