POJ 3233

思路：1.最基本的，需要用到矩阵快速幂 2.快速幂求完之后怎样快速求和？若逐项累加求和必然会超时，这时需要求递推公式：（1）若n为偶数，则：S(n) = A^(n/2)*S(n/2)+s(n/2);(2)若n为奇数 S(n) = A^(n/2+1) + S(n/2)*A^(n/2+1) + S(n/2),公式不难推，写几个就发现规律了。这样就把时间复杂度降下来了。

#include<cstdio>#include<string>#include<cstring>#include<algorithm>using namespace std;int n, m;typedef struct Matrix{    int m[30][30];    Matrix(){        memset(m, 0, sizeof(m));    }}Matrix;Matrix mtAdd(Matrix A, Matrix B){    for(int i = 0;i < n;i ++)        for(int j = 0;j < n;j ++){            A.m[i][j] += B.m[i][j];            A.m[i][j] %= m;        }    return A;}Matrix mtMul(Matrix A, Matrix B){    Matrix tmp;    for(int i = 0;i < n;i ++)        for(int j = 0;j < n;j ++)            for(int k = 0;k < n;k ++){                tmp.m[i][j] += A.m[i][k]*B.m[k][j];                tmp.m[i][j] %= m;            }    return tmp;}Matrix mtPow(Matrix A, int k){    if(k == 1) return A;    Matrix tmp = mtPow(A, k >> 1);    Matrix res = mtMul(tmp, tmp);    if(k&1) res = mtMul(res, A);    return res;}Matrix mtSum(Matrix A, int k){    if(k == 1) return A;    Matrix tmp = mtSum(A, k/2);    if(k&1){        Matrix t = mtPow(A, k/2+1);        Matrix tmp1 = mtMul(tmp, t);        Matrix tmp2 = mtAdd(t, tmp);        return mtAdd(tmp1, tmp2);    }else return mtAdd(tmp, mtMul(mtPow(A, k/2), tmp));}int main(){    int k, tmp;    /* freopen("in.c", "r", stdin); */    while(~scanf("%d%d%d", &n, &k, &m)){        Matrix M;        for(int i = 0;i < n;i ++)            for(int j = 0;j < n;j ++){                scanf("%d", &tmp);                M.m[i][j] = tmp;            }        M = mtSum(M, k);        for(int i = 0;i < n;i ++){            for(int j = 0;j < n;j ++)                printf("%d ", M.m[i][j]);            puts("");        }    }    return 0;}