acm-Matrix Power Series

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Matrix Power Series

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述

Givena n × n matrix A anda positive integer k, find thesum S = A + A² + A³ +… + A^k.

输入

The input contains exactly one test case. The first line of inputcontains three positive integers n (n ≤ 30), k (k ≤ 10^9) and m (m< 10^4). Then follow n lines each containing n nonnegativeintegers below 32,768, giving A’s elements in row-major order.

输出

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

样例输入

2 2 40 11 1

样例输出

1 22 3

来源

POJ Monthly

代码：

#include
#include
#include
#define MAX 31
int n,m,k;
typedef struct snode{
    intedge[MAX][MAX];
}Matrix;
Matrix map,ant,h,hh,c,d;

void mult(Matrix &a,Matrix &b,Matrix&c)  //矩阵C=A*B
{
    inti,j,k2;
   memset(h.edge,0,sizeof(h.edge));
   for(i=0;i
       for(j=0;j
           for(k2=0;k2
           {
               h.edge[i][j]+=(a.edge[i][k2]*b.edge[k2][j]); //**分开写，否则会WA
               h.edge[i][j]%=m;                             //**
           }
   for(i=0;i
       for(j=0;j
           c.edge[i][j]=h.edge[i][j];
}

Matrix KSM(Matrix a,intk)    //快速幂求矩阵A^k
{
    inti;
   memset(hh.edge,0,sizeof(hh.edge));
   for(i=0;i
       hh.edge[i][i]=1;
   while(k>=1)
   {
       if(k&1)
           mult(a,hh,hh);
       mult(a,a,a);
       k>>=1;
   }
    returnhh;
}

Matrix Sum(Matrix a,Matrix b)  //A,B矩阵相加
{
    inti,j;
   for(i=0;i
       for(j=0;j
       {
           h.edge[i][j]=(a.edge[i][j]+b.edge[i][j]);
           h.edge[i][j]%=m;
       }
    returnh;
}

Matrix F(int x)
{
   if(x<=1)
       return map;
    elseif(x%2) //n为奇数:F[n]=F[n/2]+F[n/2]*A^(n/2)
   {
       c=F(x-1);
       d=KSM(map,x);
       return Sum(c,d);
   }
   else         //n为偶数:F[n]=F[n-1]+A^n
   {
       c=F(x/2);
       d=KSM(map,x/2);  //二分减少规模
       mult(c,d,d);
       return Sum(c,d);
   }
}

int main()
{
    inti,j;
   scanf("%d%d%d",&n,&k,&m);
   for(i=0;i
       for(j=0;j
           scanf("%d",&map.edge[i][j]);
   ant=F(k);
   for(i=0;i
   {
       for(j=0;j
       {
           printf("%d",ant.edge[i][j]);
           if(j!=n-1)               //每行最后的空格去掉，防PE
               printf(" ");
       }
       printf("\n");
   }
    return0;
}