题解 | #买面包#

感觉题意挺意识流。所以解法也应该是意识流解法。

首先根据题意，手上有 $aa$ 元钱，每 $bb$ 元钱买一个面包，所以原价能买 $\lfloor {\displaystyle \frac{a}{b}}\rfloor \backslash left\backslash lfloor\backslash dfrac\{a\}\{b\}\backslash right\backslash rfloor$ 个面包。

然后是原价购买的面包中，每 $cc$ 个面包能换 $dd$ 个面包，所以能换到的面包数是 $\lfloor {\displaystyle \frac{\lfloor {\displaystyle \frac{a}{b}}\rfloor }{c}}\rfloor \times d\backslash left\backslash lfloor\backslash dfrac\{\backslash left\backslash lfloor\backslash dfrac\{a\}\{b\}\backslash right\backslash rfloor\}\{c\}\backslash right\backslash rfloor\backslash times\; d$ 个面包。

代码如下：

#include<cstdio>
#define int long long
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');
}
signed main(){
	int T = init();
	while (T--) {
		int a = init(), b = init(), c = init(), d = init();
		int buy = a / b;
		int ans = buy + buy / c * d;
		print(ans), putchar('\n');
	}
}