POJ 3233 Matrix Power Series 矩阵快速幂求A+A2+A3+…+Ak

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题意：给出n k m 和一个n*n的矩阵A 求A + A² +A³ + … +A^k

参考http://blog.csdn.net/wangjian8006/article/details/7868864

构造矩阵很重要啊！！！弱菜不会啊

#include <cstdio>#include <cstring>const int mod = 10000;const int maxn = 66;struct Mat{    int a[maxn][maxn];};Mat A, B;int n, m;Mat get(Mat x, Mat y){    Mat z;    memset(z.a, 0, sizeof(z.a));    for(int i = 1; i <= 2*n; i++)        for(int j = 1; j <= 2*n; j++)            for(int k = 1; k <= 2*n; k++)            {                z.a[i][j] += x.a[i][k]*y.a[k][j];                z.a[i][j] %= m;            }    return z;}void Mat_pow(int x){//puts("s");if(x <= 0)return;    while(x)    {        if(x&1)            B = get(B, A);        A = get(A, A);        x >>= 1;    }}int main(){int x;while(scanf("%d %d %d", &n, &x, &m) != EOF){memset(A.a, 0, sizeof(A.a));memset(B.a, 0, sizeof(B.a));for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)scanf("%d", &A.a[i][j]);for(int i = 1; i <= n; i++){A.a[i][i+n] = A.a[i+n][i+n] = 1;B.a[i][i] = B.a[i+n][i+n] = 1;}Mat_pow(x+1);for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(i == j){B.a[i][j+n] += m-1;B.a[i][j+n] %= m;}if(j-1)printf(" ");printf("%d", B.a[i][j+n]);}puts("");}}return 0;}

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