解題區/解題區 - UVa

UVa 12653 - Buses

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

長度為 n (保證為 5 的倍數)個對列，小公車長度為 5，種類有 K 種，大公車長度 10，種類有 L 種，請問排列的方法數有多少種？

Sample Input

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25 5 5
5 1000 1000
20 17 31
15 9 2

Sample Output

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006000
001000
111359
000765

Solution

推導公式如下

$A_{n} = K * A_{n-1} + L * A_{n-2}$

考慮塞入最後一台車的類型，找到遞迴公式。之後將其變成線性變換的結果。

$$M = \begin{bmatrix} K & 1 \\ L & 0 \end{bmatrix} \\ M^n = \begin{bmatrix} A^n & A^{n-1} \\ A^{n-1} & A^{n-2} \end{bmatrix} \\$$ 
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#include <stdio.h>
#include <string.h>
const long long mod = 1000000LL;
struct Matrix {
    long long v[2][2];
    int row, col; // row x col
    Matrix(int n, int m, long long a = 0) {
        memset(v, 0, sizeof(v));
        row = n, col = m;
        for(int i = 0; i < row && i < col; i++)
            v[i][i] = a;
    }
    Matrix operator*(const Matrix& x) const {
        Matrix ret(row, x.col);
        for(int i = 0; i < row; i++)
            for(int j = 0; j < x.col; j++)
                for(int k = 0; k < col; k++)
                    ret.v[i][j] = (ret.v[i][j] + v[i][k] * x.v[k][j]%mod)%mod;
        return ret;
    }
    Matrix operator^(const long long& n) const {
        Matrix ret(row, col, 1), x = *this;
        long long y = n;
        while(y) {
            if(y&1)	ret = ret * x;
            y = y>>1, x = x * x;
        }
        return ret;
    }
};
int main() {
    long long N, K, L;
    while (scanf("%lld %lld %lld", &N, &K, &L) == 3) {
        N /= 5;
//        long long dp[120]= {};
//        dp[0] = 1;
//        for (int i = 1; i <= N; i++) {
//            dp[i] = dp[i-1] * K;
//            if (i - 2 >= 0) {
//                dp[i] += dp[i-2] * L;
//            }
//        }
        Matrix A(2, 2, 0);
        A.v[0][0] = K%mod, A.v[0][1] = 1;
        A.v[1][0] = L%mod, A.v[1][1] = 0;
        Matrix M = A ^ N;
        printf("%06lld\n", M.v[0][0]);
    }
    return 0;
}
/*
 $ a_{n} = K a_{n-1} + L a_{n-2} $
 
 25 5 5
 5 1000 1000
 20 17 31
 15 9 2
 */