题解 | #分元宵#

$aa$ 种馅，$bb$ 种皮，也就是说每一种元宵有 $a\times ba\backslash times\; b$ 种设计方法。

一共有 $cc$ 个桌子 $dd$ 个碗，所以有 $c\times dc\backslash times\; d$ 个碗可以放元宵。

每个碗有 $a\times ba\backslash times\; b$ 种方法，所以这 $c\times dc\backslash times\; d$ 个碗：

$(a\times b)\times (a\times b)\times \cdots \times (a\times b)(a\backslash times\; b)\backslash times(a\backslash times\; b)\backslash times\backslash cdots\backslash times(a\backslash times\; b)$

一共是 $(a\times b{)}^{(c\times d)}(a\backslash times\; b)^\{(c\backslash times\; d)\}$ 种摆法，快速幂处理即可。

#include<cstdio>
#define int __int128
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 Mod;
int quick_mod(int x, int y){
    x %= Mod;
    int s = 1;
    while (y) {
        if (y & 1) s = s * x % Mod;
        x = x * x % Mod; y >>= 1;
    }
    return s;
}
signed main(){
    int a = init(), b = init(), c = init(), d = init(); Mod = init();
    a %= Mod, b %= Mod;
    if (a == 0 || b == 0) print(0), putchar('\n');
    else print(quick_mod(a * b, c * d)), putchar('\n');
}