POJ 3233

Description

Given a n × n matrix A and a positive integer k, find the sum S=A+A2+A3+…+Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

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Solution
题目大意：给一个n*n矩阵A 和 系数k， 求 S=A+A2+A3+…+Ak 。
定义一个2n*2n的矩阵B，I为单位矩阵

Bn+1=∣∣∣A0II∣∣∣n+1=∣∣∣An0A+A2+A3+...+An+II∣∣∣

求Bn+1取B[0][1]−I即可。

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Code

include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;#define maxn 100typedef long long ll;struct Mat{    int mat[maxn][maxn];//矩阵     int row, col;//矩阵行列数 };Mat mod_mul(Mat a, Mat b, int p)//矩阵乘法 {    Mat ans;    ans.row = a.row;    ans.col = b.col;    memset(ans.mat, 0, sizeof(ans.mat));    for (int i = 1;i <= ans.row;i++)        for (int j = 1;j <= ans.col;j++)            for (int k = 1;k <= a.col;k++)            {                ans.mat[i][j] += a.mat[i][k] * b.mat[k][j];                ans.mat[i][j] %= p;            }    return ans;}Mat mod_pow(Mat a, int k, int p)//矩阵快速幂 {    Mat ans;    ans.row = a.row;    ans.col = a.col;    for (int i = 1;i <= a.row;i++)        for (int j = 1;j <= a.col;j++)            ans.mat[i][j] = (i == j);    while (k)    {        if (k & 1)ans = mod_mul(ans, a, p);        a = mod_mul(a, a, p);        k >>= 1;    }    return ans;}int main(){    int n, m, k;    while (~scanf("%d%d%d", &n, &k, &m))    {        Mat A;        A.row = A.col = 2 * n;        memset(A.mat, 0, sizeof(A.mat));        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)                scanf("%d", &A.mat[i][j]);        for (int i = 1; i <= n; i++)            A.mat[i][i + n] = A.mat[i + n][i + n] = 1;//初始化单位矩阵        Mat B = mod_pow(A, k + 1, m);        for (int i = 1; i <= n; i++)            for (int j = n+1; j <= 2*n; j++)            {                if (i + n == j) printf("%d", ((B.mat[i][j] - 1) % m+m)%m);//((B.mat[i][j] - 1) % m+m)%m 避免结果为负数                else printf("%d", B.mat[i][j]);                printf("%c", j == 2 * n ? '\n' : ' ');            }    }    return 0;}