我是想先跑一边快速幂,最后乘上an-1和an-2,结果只有20分。有dalao能帮看一下么?
#include <bits/stdc++.h>
using namespace std;
#define limit (10 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(int i = a; i <= b ; ++i)
#define per(i, a, b) for(int i = b ; i >= a ; --i)
#define mint(a,b,c) min(min(a,b), c)
#define MOD 998244353
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = x * 10 + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int n ,k;
int mod;
template <int T>
struct matrix {
ll mat[T + 1][T + 1];
matrix() {
memset(mat, 0 , sizeof(mat));
}
ll mul(const matrix<T> &rhs, int x, int y) {
ll val = 0;
rep(i, 1, T) {
(val += ((mat[x][i] * rhs.mat[i][y]) % mod)) %= mod;
}
return val % mod;
}
matrix<T> operator*(const matrix<T> &rhs) {
matrix res;
rep(i, 1, T) {
rep(j, 1, T) {
res.mat[i][j] = mul(rhs, i, j);
}
}
return res;
}
static matrix<T> norm() {//单位向量
matrix ans;
rep(i ,1, T)ans.mat[i][i] = 1;
return ans;
}
void set(int i, int j, ll val){
mat[i][j] = val;
}
};
template <int T>
matrix<T> quickPow(matrix<T> base, ll expo){
matrix ans = matrix<T>::norm();//构造单位矩阵
while (expo){
if(expo & 1 )ans = ans * base;
base = base * base;
expo >>= 1;
}
return ans;
}
int p,q,first,second;
int main() {
#ifdef LOCAL
FOPEN;
#endif
matrix<2> ans;
p = read(), q = read(), first = read(), second = read(),k =read(), mod = read();
if(k < 3){
if(k == 1)write(first);
else write(second);
return 0;
}
ans.set(1,1,p);
ans.set(1,2,q);
ans.set(2,1,1);
matrix res = quickPow(ans, k - 2);//少做两次
ll ret = res.mat[1][1] * first + res.mat[1][2] * second;
ret %= mod;
write(ret);
return 0;
}