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Description

有一個隨機數列\(\{X_n\}\)，其中\(X_{n+1}=(aX_n+c)\bmod{m}\)，求\(X_n\bmod{g}\)。\((n,m,a,c,X_0\le{10^{18}},1\le{g}\le{10^8})\)

Solution

有矩陣快速冪的做法，不過也可以直接推式子。

易知\(X_n=a^nX_0+c(1+a+a^2+...+a^{n-1})\)。但是千萬不能用等比數列求和，因為這樣會產生逆元，而這題沒有保證\(gcd(a,m)=1\)。

所以可以遞迴求解等比數列之和。設\(sum(n)=\sum\limits_{i=0}^{n-1}a^i\)

當\(n\)為偶數時，\(\displaystyle \sum\limits_{i=0}^{n-1}a^i=\sum\limits_{i=0}^{\frac{n}{2}-1}a^i+\sum\limits_{i=\frac{n}{2}}^{n-1}a^i=(1+a^{\frac{n}{2}})\sum\limits_{i=0}^{\frac{n}{2}-1}a^i\)。

當\(n\)為奇數時，\(\displaystyle \sum\limits_{i=0}^{n-1}a^i=\sum\limits_{i=0}^{\frac{n-1}{2}-1}a^i+\sum\limits_{i=\frac{n-1}{2}}^{n-2}a^i+a^{n-1}=(1+a^{\frac{n-1}{2}})\sum\limits_{i=0}^{\frac{n-1}{2}-1}a^i+a^{n-1}\)

邊界：\(sum(1)=1\)。

要用快速乘，防止爆\(long\ long\)。

Code

#include <bits/stdc++.h>

using namespace std;

#define ll long long

ll m, a, c, x0, n, g;

ll read()
{
	ll x = 0ll, fl = 1ll; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') fl = -1ll; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = (x << 1ll) + (x << 3ll) + ch - '0'; ch = getchar();}
	return x * fl;
}

ll mul(ll x, ll y)
{
	ll rs = 0ll;
	while (y)
	{
		if (y & 1ll) rs = (rs + x) % m;
		x = (x + x) % m;
		y >>= 1ll;
	}
	return rs;
}

ll qpow(ll base, ll p)
{
	ll rs = 1;
	while (p)
	{
		if (p & 1ll) rs = mul(rs, base);
		base = mul(base, base);
		p >>= 1ll;
	}
	return rs;
}

ll calc(ll t)
{
	if (t == 0ll) return 0ll;
	if (t == 1ll) return 1ll;
	ll sum = 0;
	if (t % 2ll == 0ll) sum = (sum + mul(calc(t / 2ll), (1ll + qpow(a, t / 2ll))) % m) % m;
	else sum = (sum + qpow(a, t - 1ll) + mul(calc((t - 1ll) / 2ll), (1ll + qpow(a, (t - 1ll) / 2ll))) % m) % m;
	return sum % m;
}

int main()
{
	m = read(); a = read(); c = read(); x0 = read(); n = read(); g = read();
	c %= m; x0 %= m; a %= m;
	printf("%lld\n", ((mul(qpow(a, n), x0) + mul(c, calc(n))) % m + g) % g);
	return 0;
}