标签：std include 乘法 pos ace 转化 统计 fine 直接
题意：设f(n) = c ^ (2n - 6) * f(n - 1) * f(n - 2) * f(n - 3), 问第n项是多少？
思路：官方题解：我们先转化一下，令g(x) = c ^ x * f(x), 那么原式转化为了g(x) = g(x - 1) * g(x - 2) * g(x - 3)。之后我们可以考虑把f(1), f(2), f(3)和c的质因子找出来，枚举质因子对答案的贡献。我们发现，如果是质因子的数目的话，乘法就变成了加法（相当于统计质因子的指数），这样就可以用矩阵乘法优化了。注意，矩阵转移的时候，模数是1e9 + 6，因为转移的时候是指数（欧拉定理）。其实基于这种想法，我们可以不用处理质因子，直接计算g(1), g(2)，g(3)对答案的贡献。
代码：
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod = 1e9 + 7;
const LL mod1 = 1e9 + 6;
map<LL, LL> mp[4];
set<LL> s;
set<LL>::iterator it;
struct Matrix {
LL a[3][3];
void init(LL num = 0) {
memset(a, 0, sizeof(a));
for (int i = 0; i < 3; i++)
a[i][i] = num;
}
Matrix operator * (const Matrix& now) const {
Matrix res;
res.init();
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
for (int k = 0; k < 3; k++)
res.a[i][j] = (res.a[i][j] + (a[i][k] * now.a[k][j]) % mod1) % mod1;
return res;
}
Matrix operator ^ (const LL num) const {
Matrix ans, x = *this;
ans.init(1);
LL now = num;
for (; now; now >>= 1) {
if(now & 1) ans = ans * x;
x = x * x;
}
return ans;
}
void print() {
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
printf("%lld ", a[i][j]);
}
printf("\n");
}
}
};
void div(LL num, int pos) {
for (LL i = 2; i * i <= num; i++) {
if(num % i == 0) {
s.insert(i);
while(num % i == 0) {
mp[pos][i]++;
num /= i;
}
}
}
if(num > 1) {
s.insert(num);
mp[pos][num]++;
}
}
LL qpow(LL x, LL y) {
LL ans = 1;
for (; y; y >>= 1ll) {
if(y & 1ll) ans = (ans * x) % mod;
x = (x * x) % mod;
}
return ans;
}
LL a[4];
int main() {
LL n;
scanf("%lld", &n);
for (int i = 0; i < 4; i++) {
scanf("%lld", &a[i]);
div(a[i], i);
}
Matrix x, y, x1;
x.init();
x.a[0][2] = x.a[1][2] = x.a[2][2] = x.a[2][1] = x.a[1][0] = 1;
x = x ^ (n - 1);
LL ans = 1;
for (it = s.begin(); it != s.end(); it++) {
y.init();
for (int i = 0; i < 3; i++) {
y.a[0][i] = mp[i][*it];
}
for (int i = 0; i < 3; i++)
y.a[0][i] = (y.a[0][i] + ((LL)(i + 1) * mp[3][*it] % mod)) % mod;
y = y * x;
ans = (ans * qpow(*it, y.a[0][0]) % mod) % mod;
}
ans = ans * qpow(qpow(a[3], mod - 2), n) % mod;
printf("%lld\n", ans);
}
标签：std include 乘法 pos ace 转化 统计 fine 直接
原文地址：https://www.cnblogs.com/pkgunboat/p/11012730.html
