POJ 3233 —— 矩阵快速幂

Matrix Power Series Time Limit: 3000MS   Memory Limit: 131072K Total Submissions: 12857   Accepted: 5523 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 nnonnegative 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 Source POJ Monthly--2007.06.03, Huang, Jinsong

题意是给你一个n * n的矩阵A和k ， m 。S = A + A ^ 2 + A ^ 3 + …… A ^ k 对m取模后的结果。

显然不能直接去矩阵快速幂，我们构建一个矩阵{{A , 0 } , {I , I}};

可以算出{ {A^k} , {S} } = { {A , 0} , {I , I} } ^ k * { {I} , {0} };

没有画图工具只能文字描述了。。。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 50 + 5;
const int MAXS = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-8
const long long MOD = 1000000000 + 7;
const int mod = 10007;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
typedef vector<int> vec;
typedef vector<vec> mat;


#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int n  , m;
int k;
mat A;
mat mul(mat & A , mat & B)
{
    mat C(A.size() , vec(B[0].size()));
    for(int i = 0 ; i < A.size() ; i++)
    {
        for(int k = 0 ; k < B.size() ; k++)
        {
            for(int j = 0  ; j < B[0].size() ; j++)
            {
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % m;
            }
        }
    }
    return C;
}
mat pow(mat A , LL n)
{
    mat B(A.size() , vec(A.size()));
    for(int i = 0 ; i < A.size() ; i++)B[i][i] = 1;
    while(n > 0)
    {
        if(n & 1)B = mul(B , A);
        A = mul(A , A);
        n >>= 1;
    }
    return B;
}
void solve()
{
    mat B(n * 2 , vec(n * 2));
    FOR(i , 0 , n)
    {
        FOR(j , 0 , n)
        {
            B[i][j] = A[i][j];
        }
        B[n + i][i] = B[n + i][n + i] = 1;
    }
    B = pow(B , k + 1);
    FOR(i , 0 , n)
    {
        FOR(j ,0 , n)
        {
            int a = B[n + i][j] % m;
            if(i == j)a = (a + m - 1) % m;
            printf("%d%c" , a , j == n - 1 ? '\n' : ' ');
        }
    }
}

int main()
{
    while(~scanf("%d%d%d" , &n , &k , &m))
    {
        int xx;
        A.clear();
//        FOR(i , 0 , n)A[i].clear();
        FOR(i , 0 , n)
        {
            vec x;
            FOR(j , 0, n)
            {
                scanf("%d" , &xx);
                x.push_back(xx);
            }
            A.push_back(x);
        }

        solve();
    }
    return 0;
}