[Luogu] P2044 [NOI2012]随机数生成器
Description
有一个随机数列\(\{X_n\}\),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
有矩阵快速幂的做法,不过也可以直接推式子。
易知\(X_n=a^nX_0+c(1+a+a^2+...+a^{n-1})\)。但是千万不能用等比数列求和,因为这样会产生逆元,而这题没有保证\(gcd(a,m)=1\)。
所以可以递归求解等比数列之和。设\(sum(n)=\sum\limits_{i=0}^{n-1}a^i\),分两种情况讨论:
当\(n\)为偶数时,\(\displaystyle \sum\limits_{i=0}^{n-1}a^i=\sum\limits_{i=0}^{\frac{n}{2}-1}a^i+\sum\limits_{i=\frac{n}{2}}^{n-1}a^i=(1+a^{\frac{n}{2}})\sum\limits_{i=0}^{\frac{n}{2}-1}a^i\)。
当\(n\)为奇数时,\(\displaystyle \sum\limits_{i=0}^{n-1}a^i=\sum\limits_{i=0}^{\frac{n-1}{2}-1}a^i+\sum\limits_{i=\frac{n-1}{2}}^{n-2}a^i+a^{n-1}=(1+a^{\frac{n-1}{2}})\sum\limits_{i=0}^{\frac{n-1}{2}-1}a^i+a^{n-1}\)。
边界:\(sum(1)=1\)。
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;
}