POJ 3233 Matrix Power Series 矩阵等比数列求和
和前面有一题是一样的做法吧。
A^1+A^2+A^3+A^4 = A^1+A^2+A^2*(A^1+A^2)类似这样搞就可以二分处理了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>
using namespace std;
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 40;
LL N, K, MOD;
struct Matrix {
int n, m;
LL data[maxn][maxn];
Matrix(int n = 0, int m = 0): n(n), m(m) {
memset(data, 0, sizeof(data));
}
};
Matrix operator * (Matrix a, Matrix b) {
Matrix ret(a.n, b.m);
for(int i = 1; i <= a.n; i++) {
for(int j = 1; j <= b.m; j++) {
for(int k = 1; k <= a.m; k++) {
ret.data[i][j] += a.data[i][k] * b.data[k][j];
ret.data[i][j] %= MOD;
}
}
}
return ret;
}
Matrix operator + (Matrix a, Matrix b) {
for(int i = 1; i <= a.n; i++) {
for(int j = 1; j <= a.m; j++) {
a.data[i][j] += b.data[i][j];
a.data[i][j] %= MOD;
}
}
return a;
}
Matrix pow(Matrix mat, LL p) {
if(p == 1) return mat;
Matrix ret = pow(mat * mat, p / 2);
if(p & 1) ret = ret * mat;
return ret;
}
Matrix calc(Matrix mat, LL K) {
if(K == 1) return mat;
Matrix ret = calc(mat, K / 2);
ret = ret + pow(mat, K / 2) * ret;
if(K & 1) ret = ret + pow(mat, K);
return ret;
}
int main() {
while(scanf("%lld%lld%lld", &N, &K, &MOD) != EOF) {
Matrix mat(N, N);
for(int i = 1; i <= N; i++) {
for(int j = 1; j <= N; j++) {
scanf("%lld", &mat.data[i][j]);
}
}
mat = calc(mat, K);
for(int i = 1; i <= N; i++) {
for(int j = 1; j <= N; j++) {
printf("%lld ", mat.data[i][j]);
}
puts("");
}
}
return 0;
}

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