POJ3233Matrix Power Series(矩阵快速幂)

题意

给出$n \times n$的矩阵$a$，求$\sum_{i = 1}^k a^i $，每个元素对$m$取模

sol

考虑直接分治

当$k$为奇数时

$\sum_{i = 1}^k a^i = \sum_{i = 1}^{k / 2 + 1} a^i + a^{k / 2 + 1}(\sum_{i = 1}^{k / 2} a^i)$

当$k$为偶数时

$sum_{i = 1}^k = \sum_{i = 1}^{k / 2} a^i + a^{k / 2}(\sum_{i = 1}^{k / 2}a^i)$

 

当然还可以按套路对前缀和构造矩阵也是可以做的。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#define ll long long 
using namespace std;
int n, k, mod;
int mul(int x, int y) {
    if(1ll * x * y > mod) return 1ll * x * y % mod;
    else return 1ll * x * y;
}
int add(int x, int y) {
    if(x + y > mod) return x + y - mod;
    else return x + y;
}
struct matrix {
    int m[31][31];
    matrix() {
        memset(m, 0, sizeof(m));
    }
    bool operator < (const matrix &rhs) const {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                if(m[i][j] != rhs.m[i][j])
                    return m[i][j] < rhs.m[i][j];
        return 1;
    }
    matrix operator * (const matrix &rhs) const {
        matrix ans;
        for(int k = 1; k <= n; k++)
            for(int i = 1; i <= n; i++)
                for(int j = 1; j <= n; j++)
                    ans.m[i][j] = add(ans.m[i][j], mul(m[i][k], rhs.m[k][j]));
        return ans;
    }
    matrix operator + (const matrix &rhs) const {
        matrix ans;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                ans.m[i][j] = add(m[i][j], rhs.m[i][j]);
        return ans;
    }
}a;
matrix getbase() {
    matrix base;
    for(int i = 1; i <= n; i++) base.m[i][i] = 1;
    return base;
}
matrix fp(matrix a, int p) {
    matrix base = getbase();
    while(p) {
        if(p & 1) base = base * a;
        a = a * a; p >>= 1;
    }
    return base;
}
matrix solve(int k) {
    if(k == 1) return a;
    matrix res = solve(k / 2);
    if(k & 1) {
        matrix po = fp(a, k / 2 + 1);
        return res + po + po * res;
    }
    else return res + fp(a, k / 2) * res;

}
main() {
//    freopen("a.in", "r", stdin);
    cin >> n >> k >> mod;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            cin >> a.m[i][j];
    matrix ans = solve(k);
    for(int i = 1; i <= n; i++, puts(""))
        for(int j = 1; j <= n; j++)
            printf("%d ", ans.m[i][j] % mod);
}