题解 | #简单题#

作为小白我们虽然看不懂第一个式子到底等于什么，但是也可以做这道题。

考虑第二个式子和第三个式子，实际上就可以化成 $\beta \times {\theta }^{\alpha }\backslash beta\backslash times\backslash theta^\{\backslash alpha\}$ 的形式。

然后把样例的两个值代入，

$\{\begin{array}{c}196\times {\theta }^5=29089.0\\ 60\times {\theta }^3=1205.1322\end{array}\backslash begin\{cases\}196\backslash times\backslash theta^5=29089.0\backslash \backslash 60\backslash times\backslash theta^3=1205.1322\backslash end\{cases\}$

解得 $\theta \backslash theta$ 约为 $2.71828\cdots 2.71828\backslash cdots$（用第二个式子计算 $\theta ={\left({\displaystyle \frac{1205.1322}{60}}\right)}^{{\displaystyle \frac{1}{3}}}\backslash theta=\backslash left(\backslash dfrac\{1205.1322\}\{60\}\backslash right)^\{\backslash dfrac\{1\}\{3\}\}$ 即可）

这很明显就反应出来是这个东西：$ee$。

然后上网找一下 $ee$ 的前 $1010$ 几位，复制进代码即可：$e=2.7182818284590452\cdots e=2.7182818284590452\backslash cdots$

#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", beta * pow(theta, alpha)); break; }
			case 2 : { printf("%.2lf\n", beta * pow(theta, alpha)); break; }
			case 3 : { printf("%.3lf\n", beta * pow(theta, alpha)); break; }
			case 4 : { printf("%.4lf\n", beta * pow(theta, alpha)); break; }
			case 5 : { printf("%.5lf\n", beta * pow(theta, alpha)); break; }
		}
	}
}