• POJ 3233 Matrix Power Series 矩阵等比数列求和


    和前面有一题是一样的做法吧。

    A^1+A^2+A^3+A^4 = A^1+A^2+A^2*(A^1+A^2)类似这样搞就可以二分处理了。

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <climits>
    #include <iostream>
    #include <string>
    
    using namespace std;
     
    #define MP make_pair
    #define PB push_back
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    typedef pair<double, double> PDD;
    const int INF = INT_MAX / 3;
    const double eps = 1e-8;
    const LL LINF = 1e17;
    const double DINF = 1e60;
    const int maxn = 40;
    LL N, K, MOD;
    
    struct Matrix {
        int n, m;
        LL data[maxn][maxn];
        Matrix(int n = 0, int m = 0): n(n), m(m) {
            memset(data, 0, sizeof(data));
        }
    };
    
    Matrix operator * (Matrix a, Matrix b) {
        Matrix ret(a.n, b.m);
        for(int i = 1; i <= a.n; i++) {
            for(int j = 1; j <= b.m; j++) {
                for(int k = 1; k <= a.m; k++) {
                    ret.data[i][j] += a.data[i][k] * b.data[k][j];
                    ret.data[i][j] %= MOD;
                }
            }
        }
        return ret;
    }
    
    Matrix operator + (Matrix a, Matrix b) {
        for(int i = 1; i <= a.n; i++) {
            for(int j = 1; j <= a.m; j++) {
                a.data[i][j] += b.data[i][j];
                a.data[i][j] %= MOD;
            }
        }
        return a;
    }
    
    Matrix pow(Matrix mat, LL p) {
        if(p == 1) return mat;
        Matrix ret = pow(mat * mat, p / 2);
        if(p & 1) ret = ret * mat;
        return ret;
    }
    
    Matrix calc(Matrix mat, LL K) {
        if(K == 1) return mat;
        Matrix ret = calc(mat, K / 2);
        ret = ret + pow(mat, K / 2) * ret;
        if(K & 1) ret = ret + pow(mat, K);
        return ret;
    }
    
    int main() {
        while(scanf("%lld%lld%lld", &N, &K, &MOD) != EOF) {
            Matrix mat(N, N);
            for(int i = 1; i <= N; i++) {
                for(int j = 1; j <= N; j++) {
                    scanf("%lld", &mat.data[i][j]);
                }
            }
            mat = calc(mat, K);
            for(int i = 1; i <= N; i++) {
                for(int j = 1; j <= N; j++) {
                    printf("%lld ", mat.data[i][j]);
                }
                puts("");
            }
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/4049038.html
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