嘟嘟嘟
这题就是一道矩阵加速dp的水题，dp式都给你了，所以矩阵这方面就不说了。
之所以发这篇博客，是因为两数相乘可能会爆long long，所以得用快速乘。
现学了一下，感觉和快速幂特别像。
对于两个数\(a, b\)，按位枚举\(b\)，如果\(b\)的第\(i\)位为\(1\)，答案就加上\(a * 2 ^ i\)。
发一下代码。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
//const int maxn = ;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) {last = ch; ch = getchar();}
while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
ll mod, a, c, x0, n, g;
ll mul(ll a, ll b) //神奇的快速乘
{
ll ret = 0;
for(; b; b >>= 1, a = (a << 1) % mod)
if(b & 1) ret = (ret + a) % mod;
return ret;
}
const int N = 2;
struct Mat
{
ll a[N][N];
Mat operator * (const Mat& oth)const
{
Mat ret; Mem(ret.a, 0);
for(int i = 0; i < N; ++i)
for(int j = 0; j < N; ++j)
for(int k = 0; k < N; ++k)
ret.a[i][j] += mul(a[i][k], oth.a[k][j]), ret.a[i][j] %= mod;
return ret;
}
}F;
void init()
{
Mem(F.a, 0);
F.a[0][0] = a; F.a[0][1] = F.a[1][1] = 1;
}
Mat quickpow(Mat A, ll b)
{
Mat ret; Mem(ret.a, 0);
for(int i = 0; i < N; ++i) ret.a[i][i] = 1;
for(; b; b >>= 1, A = A * A)
if(b & 1) ret = ret * A;
return ret;
}
int main()
{
mod = read(), a = read(), c = read(), x0 = read(), n = read(), g = read();
a %= mod; c %= mod; x0 %= mod;
init();
Mat A = quickpow(F, n);
write((mul(A.a[0][0], x0) + mul(A.a[0][1], c)) % mod % g), enter;
return 0;
}
