POJ 3233 Matrix Power Series（矩阵快速幂）

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
思路:当k=6时可以将S化解成如下等式进行二分求解

S=（A+A2+A3）+A3∗（A+A2+A3）

#include <stdio.h>#include <algorithm>using namespace std;const int MAX_N=35;int inf=99999999;typedef int Matrix[MAX_N][MAX_N];int n,m;Matrix A;void MATRIX_MUL(const Matrix p,const Matrix q,Matrix aim)//矩阵乘法{    for(int i=0;i<n;i++)    {        for (int j=0;j<n;j++)        {            aim[i][j]=0;            for (int k=0;k<n;k++)                aim[i][j]+=p[i][k]*q[k][j];            aim[i][j]%=m;        }    }}void MATRIX_PLU(const Matrix p,const Matrix q,Matrix aim)//矩阵加法{    for(int i=0;i<n;i++)    {        for (int j=0;j<n;j++)        {            aim[i][j]=p[i][j]+q[i][j];            aim[i][j]%=m;        }    }}void MATRIX_POW(int k,Matrix aim)//求解矩阵快速幂，将结果保存在aim里{    if(k==1)    {        for (int i=0;i<n;i++)            for (int j=0;j<n;j++)                aim[i][j]=A[i][j];    }    else    {        if(k%2==1)        {            Matrix l,r;            MATRIX_POW(k/2,l);            MATRIX_MUL(l,l,r);            MATRIX_MUL(r,A,aim);        }        else        {            Matrix a;            MATRIX_POW(k/2,a);            MATRIX_MUL(a,a,aim);        }    }}void solve(int k,Matrix aim)//求解一个值为k的矩阵和式，将结果保存在aim里{    if(k==1)    {        for (int i=0;i<n;i++)        {            for (int j=0;j<n;j++)                aim[i][j]=A[i][j];        }    }    else    {        if(k%2==1)        {            Matrix a1,a2,a3,ak;            solve(k/2,a1);            MATRIX_POW(k/2,a3);            MATRIX_MUL(a3,A,ak);            MATRIX_PLU(a1,ak,a2);            MATRIX_MUL(a2,a3,ak);            MATRIX_PLU(a1,ak,aim);        }        else        {            Matrix a1,a2,ap;            solve(k/2,a1);            MATRIX_POW(k/2,ap);            MATRIX_MUL(a1,ap,a2);            MATRIX_PLU(a1,a2,aim);        }    }}int main(){    freopen("test.txt","r",stdin);    int k;    scanf("%d%d%d",&n,&k,&m);    for (int i=0;i<n;i++)    {        for (int j=0;j<n;j++)        {            scanf("%d",&A[i][j]);            A[i][j]%=m;        }    }    Matrix ans;    solve(k,ans);    for (int i=0;i<n;i++)    {        for (int j=0;j<n;j++)            printf("%d ",ans[i][j]);        printf("\n");    }    return 0;}