POJ 3233 Matrix Power Series（矩阵快速幂）

Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 19338 Accepted: 8161
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

可以找到递推关系 : s[k]=s[k-1]+A^k;
然后构造矩阵，利用矩阵快速幂

具体见代码

#include <iostream>#include <string.h>#include <algorithm>#include <math.h>#include <stdio.h>#include <stdlib.h>using namespace std;int n,k;int m;struct Node{    int a[65][65];};Node multiply(Node a,Node b){    Node c;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            c.a[i][j]=0;            for(int k=1;k<=n;k++)            {                (c.a[i][j]+=(a.a[i][k]*b.a[k][j])%m)%=m;             }        }    }     return c;}Node quick(Node a,int x){    Node c;    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        c.a[i][j]=(i==j?1:0);    for(x;x>0;x>>=1)    {        if(x&1)             c=multiply(c,a);        a=multiply(a,a);     }    return c;}int main(){   while( scanf("%d%d%d",&n,&k,&m)!=EOF)   {    Node a;Node b;Node c;    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+n]);            b.a[i+n][j+n]=a.a[i][j+n];        }    for(int i=1;i<=n;i++)    {        b.a[i][i]=1;        b.a[i+n][i]=1;    }    n=n*2;    c=multiply(a,quick(b,k));    for(int i=1;i<=n/2;i++)        for(int j=1;j<=n/2;j++)           if(j==n/2)printf("%d\n",c.a[i][j]);           else printf("%d ",c.a[i][j]);   }    return 0;}