題目描述

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 32768, 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

題目分析

首先我們可以發現我們需要的就是每個矩陣的和那麼我們用樣例舉例子

程式碼

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 53;
int MOD;
struct Matrix {
    int Ma[MAXN+10][MAXN+10];
    int n, m;
    void Clear(int u, int un, int um){
        n = un, m = um;
        for(int i=1;i<=un;i++)
            for
 
(int j=1;j<=um;j++)
                Ma[i][j] = 0;
        if(u) for(int i=1;i<=un;i++)
            Ma[i][i] = 1;
    }
    Matrix operator* (const Matrix& ma) {
        Matrix ret ;
        ret.Clear(0, n, m);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=ma.m;j++){
                for(int k=1;k<=ma.n;k++){
                    ret.Ma[i][j] += Ma[i][k] * ma.Ma[k][j];
                    ret.Ma[i][j] %= MOD;
                }
            }
        }
        return ret;
    }
    void Print(){
        for(int i=1;i<=n;i++){
            printf("%d", Ma[i][1]);
            for(int j=2;j<=m;j++)
                printf(" %d", Ma[i][j]);
            printf("\n");
        }
    }
};
Matrix Mpow(Matrix m, int p){
    Matrix ret;
    if(p == 0){
        ret.Clear(1, 2, 2);
        return ret;
    }else if(p == 1) return m;
    ret = Mpow(m, p/2);
    if(p%2 == 0) return ret * ret;
    return (ret * ret) * m;
}
int main(){
    int n, k;
    scanf("%d%d%d", &n, &k, &MOD);
    Matrix m, pm;
    m.Clear(0, n*2, n*2);
    pm.Clear(0, n*2, n*2);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d", &m.Ma[i][j]);
            pm.Ma[i][j] = m.Ma[i][j];
        }
        pm.Ma[i][i+n] = 1;
        pm.Ma[i+n][i+n] = 1;
    }
    //pm.Print();
    Matrix ans = Mpow(pm, k);
    Matrix tans = m * ans;
    for(int i=1;i<=n;i++){
        printf("%d", tans.Ma[i][1+n]);
        for(int j=2;j<=n;j++)
            printf(" %d", tans.Ma[i][j+n]);
        printf("\n");
    }

    return 0;
}