題意就是一個等比數列求和的意思，只不過每一項都是矩陣

這裡需要進行一下轉移矩陣的構造，形成一個遞推累加的效果：

設 B = (A,I;0,I)

則B^(k + 1) = (A^(k + 1),I + A + A^2 + A^3 + … + A^k;0,I)

原題可解，程式碼如下

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
 

#include <vector>
#include <stack>
#include <set>
#include <map>
using namespace std;
const int MAXN = 80;
int n, k, m;
struct  Matrix
{
    __int64 a[MAXN][MAXN];
    int r;//行數
    int c;//列數
}org, ans;
void print()
{
    for (int i = 0;i < n;i++)
    {
        for (int j = 0;j < n;j++)
        {
            if
 
 (j) cout << " ";
            if (i != j) cout << ans.a[i][j + n];
            else cout << (ans.a[i][j + n] - 1 < 0 ? m - 1 : ans.a[i][j + n] - 1);
        }
        cout << endl;
    }
}

Matrix mul(Matrix mat1, Matrix mat2)
{
    Matrix res;
    res.c = mat2.c;
    res.r = mat1.r;
    memset
 
(res.a, 0, sizeof(res.a));
    for (int i = 0;i < mat1.r;i++)
    {
        for (int j = 0;j < mat2.c;j++)
        {
            for (int k = 0;k < mat1.c;k++)
            {
                res.a[i][j] += mat1.a[i][k] * mat2.a[k][j];
                res.a[i][j] %= m;
            }
            res.a[i][j] %= m;
        }
    }
    return res;
}

void fmp()
{
    k++;
    while (k)
    {
        if (k & 1) ans = mul(ans, org);
        org = mul(org, org);
        k >>= 1;
    }
}

int main()
{
    while(cin>>n>>k>>m)
    {
        int i, j;
        memset(org.a, 0, sizeof(org.a));
        memset(ans.a, 0, sizeof(ans.a));
        for (i = 0;i<n;i++) {
            for (j = 0;j<n;j++)
                scanf("%d", &org.a[i][j]);
        }
        for (i = 0;i<n;i++) {
            org.a[i + n][i + n] = org.a[i][i + n] = 1;
            ans.a[i][i] = ans.a[i + n][i + n] = 1;
        }
        org.c = org.r = 2 * n;
        ans.c = ans.r = 2 * n;
        fmp();
        print();
    }
    return 0;
}