POJ 3233 Matrix Power Series(矩阵的快速幂)

Matrix Power Series

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 17841 Accepted: 7538

Description

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

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 40 11 1

Sample Output

1 22 3

解题思路:

分别用矩阵快速幂求出每一项A^i,然后把每一项矩阵相加即可,由于k很大，必须采用二分求解。

AC代码:

#include<iostream>#include<cstdlib>const int maxn = 101;using namespace std;typedef struct{int m[maxn][maxn];}matrax;matrax a,per;int n,M;void init(){int i,j;for(i=0;i<n;i++){ for(j=0;j<n;j++){scanf("%d",&a.m[i][j]);a.m[i][j] %= M;per.m[i][j] = (i == j); //单位矩阵}}}matrax add(matrax a,matrax b){matrax c;int i,j;for(i=0;i<n;i++){for(j=0;j<n;j++){c.m[i][j] = (a.m[i][j] + b.m[i][j]) % M;}}return c;}matrax multi(matrax a,matrax b){matrax c;int k,i,j;for(i=0;i<n;i++){for(j=0;j<n;j++){c.m[i][j] = 0;for(k=0;k<n;k++){c.m[i][j] += a.m[i][k] * b.m[k][j];}}c.m[i][j] %= M;}return c;}matrax power(int k){matrax c,p,ans = per;p = a;while(k){if(k & 1){ans = multi(ans,p);k--;}else{k /= 2;p = multi(p,p);}}return ans;}matrax Matraxsum(int k){if(k == 1)return a;matrax temp,b;temp = Matraxsum(k / 2);if(k & 1){b = power(k / 2 + 1);temp = add(temp,multi(temp,b));temp = add(temp,b);}else{b = power(k / 2);temp = add(temp,multi(temp,b));}return temp;}int main(){int k,i,j;while(scanf("%d%d%d",&n,&k,&M) != EOF){ init();matrax ans = Matraxsum(k);for(i=0;i<n;i++){for(j=0;j<n-1;j++){printf("%d ",ans.m[i][j]);}printf("%d\n",ans.m[i][j]);}}return 0;}