POJ 3233 Matrix Power Series

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

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【题目分析】
只需要分治一下，从中间分开，然后做logk次快速幂就可以了。详解见代码

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【代码】

#include <cstdio>#include <iostream>#include <cstring>using namespace std;typedef struct node{    int b[32][32];}matx;matx a,sa,unit;int n,m,k;matx add(matx a,matx b){    matx c;    for (int i=0; i<n; i++)      for (int j=0; j<n; j++)      {          c.b[i][j]=a.b[i][j]+b.b[i][j];          c.b[i][j]%=m;      }    return c;}matx mul(matx a,matx b){    matx c;    for (int i=0; i<n; i++)      for (int j=0; j<n; j++)      {          c.b[i][j]=0;          for (k=0; k<n; k++)             c.b[i][j]+=a.b[i][k]*b.b[k][j];          c.b[i][j]%=m;      }    return c;}matx cal(int exp){    matx p,q;    p=a;    q=unit;    while (exp!=1)    {        if (exp&1)        {            exp--;            q=mul(p,q);        }        else        {            exp>>=1;            p=mul(p,p);        }    }    return mul(p,q);}matx ms(int k){    if (k==1) return a;    matx tmp,tnow;    tmp=ms(k/2);    if (k&1)    //k为奇数时sum(k)=(1+A^(k/2+1))*sum(k/2)+A^(k/2+1);    {        tnow=cal(k/2+1);        tmp=add(tmp,mul(tmp,tnow));        tmp=add(tnow,tmp);    }       else        //k为偶数时sum(k)=(1+A^(k/2))*sum(k/2)    {        tnow=cal(k/2);        tmp=add(tmp,mul(tmp,tnow));    }    return tmp;}int main(){    scanf("%d%d%d",&n,&k,&m);    for (int i=0; i<n; i++)       for (int j=0; j<n; j++)       {           scanf("%d",&a.b[i][j]);           a.b[i][j]%=m;           unit.b[i][j]=(i==j);       }    sa=ms(k);    for (int i=0; i<n; i++)    {        for (int j=0; j<n; j++)printf("%d ",sa.b[i][j]);        printf("\n");    }    return 0;}