uva11149 - Power of Matrix 矩阵倍增

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原创最新推荐文章于 2026-01-28 20:20:59 发布 739 阅读 AI 写同款· GEO 优化 ›

20、啁啾脉冲光纤放大器：原理、设计与仿真本文详细介绍了啁啾脉冲光纤放大器（FCPA）的原理、设计与仿真过程。内容涵盖CPA技术的基本原理、光纤基CPA系统的组成、Tm掺杂光纤放大器的工作机制、全光纤脉冲展宽器的设计以及仿真分析。通过MATLAB仿真展示了脉冲在光纤中的展宽与放大过程，并讨论了如何优化系统以减少非线性失真、控制峰值功率并提高整体性能。文章还总结了系统优化的关键策略，并展望了未来的发展方向。阅读详情

Consider an n-by-n matrixA. We define A^k = A *A * ... * A (k times). Here, * denotes the usual matrix multiplication.

You are to write a program that computes the matrix A + A² + A³ + ... + A^k.

Example

Suppose A = . ThenA² = =, thus:

Such computation has various applications. For instance, the above example actually counts all the paths in the following graph:

Input consists of no more than 20 test cases. The first line for each case contains two positive integersn (≤ 40) and k (≤ 1000000). This is followed by n lines, each containing n non-negative integers, giving the matrixA.

Input is terminated by a case where n = 0. This case need NOT be processed.

Output

For each case, your program should compute the matrix A + A² + A³ + ... + A^k. Since the values may be very large, you only need to print theirlast digit. Print a blank line after each case.

Sample Input

Sample Output

0 2 4
0 0 2
0 0 0

计算A+A^2+A^3...+A^k。

当k是偶数，A+A^2+A^3...+A^k=（E+A^(k/2)）*(A+A^2...+A^(k/2))。当k是奇数，A+A^2+A^3...+A^k=（E+A^(k/2)）*(A+A^2...+A^(k/2))+A^k。

类似二分的思想。话说被重载运算符优先级坑了，看来以后重载运算符要打括号啊。。

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#define INF 0x3f3f3f3f
#define MAXN 45
#define MAXM 110
#define MOD 1000000007
#define MAXNODE 4*MAXN
#define eps 1e-9
using namespace std;
typedef long long LL;
int N,K;
struct Mat{
    int mat[MAXN][MAXN];
    Mat(){
        memset(mat,0,sizeof(mat));
    }
};
Mat operator * (Mat a,Mat b){
    int i,j,k;
    Mat ret;
    for(k=0;k<N;k++){
        for(i=0;i<N;i++){
            if(a.mat[i][k]==0) continue;
            for(j=0;j<N;j++){
                if(b.mat[k][j]==0) continue;
                ret.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                ret.mat[i][j]%=10;
            }
        }
    }
    return ret;
}
Mat operator + (Mat a,Mat b){
    Mat ret;
    for(int i=0;i<N;i++)
        for(int j=0;j<N;j++) ret.mat[i][j]=(a.mat[i][j]+b.mat[i][j])%10;
    return ret;
}
Mat operator ^ (Mat a,int n){
    Mat ret,t=a;
    int i,j;
    for(i=0;i<N;i++)
        for(j=0;j<N;j++) ret.mat[i][j]=(i==j);
    while(n){
        if(n&1) ret=ret*t;
        t=t*t;
        n>>=1;
    }
    return ret;
}
Mat solve(Mat a,int k){
    if(k==1) return a;
    Mat ans;
    for(int i=0;i<N;i++) ans.mat[i][i]=1;
    if(k==0) return ans;
    ans=((a^(k>>1))+ans)*solve(a,k>>1);
    if(k%2) ans=(a^k)+ans;
    return ans;
}
Mat a,ans;
int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&N,&K)!=EOF&&N){
        for(int i=0;i<N;i++)
            for(int j=0;j<N;j++){
                int t;
                scanf("%d",&t);
                a.mat[i][j]=t%10;
            }
        ans=solve(a,K);
        for(int i=0;i<N;i++){
            for(int j=0;j<N-1;j++) printf("%d ",ans.mat[i][j]);
            printf("%d\n",ans.mat[i][N-1]);
        }
        puts("");
    }
    return 0;
}