矩阵快速幂 toj 3016

3016.   Decode the Strings

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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 1559   Accepted Runs: 500

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Bruce Force has had an interesting idea how to encode strings. The following is the description of how the encoding is done:

Let x₁,x₂,...,x_n be the sequence of characters of the string to be encoded.

1. Choose an integer m and n pairwise distinct numbers p₁,p₂,...,p_n from the set {1, 2, ..., n} (a permutation of the numbers 1 to n).
2. Repeat the following step m times.
3. For 1 ≤ i ≤ n set y_i to x_pi, and then for 1 ≤ i ≤ n replace x_i by y_i.

For example, when we want to encode the string "hello", and we choose the value m = 3 and the permutation 2, 3, 1, 5, 4, the data would be encoded in 3 steps: "hello" -> "elhol" -> "lhelo" -> "helol".

Bruce gives you the encoded strings, and the numbers m and p₁, ..., p_n used to encode these strings. He claims that because he used huge numbers m for encoding, you will need a lot of time to decode the strings. Can you disprove this claim by quickly decoding the strings?

Input

The input contains several test cases. Each test case starts with a line containing two numbers n and m (1 ≤ n ≤ 80, 1 ≤ m ≤ 10⁹). The following line consists of n pairwise different numbers p₁,...,p_n (1 ≤ p_i ≤ n). The third line of each test case consists of exactly n characters, and represent the encoded string. The last test case is followed by a line containing two zeros.

Output

For each test case, print one line with the decoded string.

Sample Input

5 3
2 3 1 5 4
helol
16 804289384
13 10 2 7 8 1 16 12 15 6 5 14 3 4 11 9
scssoet tcaede n
8 12
5 3 4 2 1 8 6 7
encoded?
0 0

Sample Output

hello
second test case
encoded?

 解析：1、置换矩阵

           2、置换矩阵逆矩阵求法

           3、矩阵快速幂

 

代码：

#include <iostream>
#include <cstring>
using namespace std;
const int maxN = 85;

int A[maxN][maxN],B[maxN][maxN],res[maxN][maxN];
int n,m;
void arrayMul(int A[maxN][maxN],int B[maxN][maxN])
{
    int C[maxN][maxN] = {0};
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            for(int k=0;k<n;k++)
                C[i][j] += A[i][k]*B[k][j];
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            A[i][j] = C[i][j];
}
void arrayPow(int A[maxN][maxN],int k)
{
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            res[i][j] = (i == j);
    while(k > 0)
    {
        if(k & 1) arrayMul(res,A);
        k >>= 1;
        arrayMul(A,A);
    }
}
int main()
{
    int t;
    string key;
    while(cin >> n >> m, m + n)
    {
        memset(A,0,sizeof(A));
        for(int i=0;i<n;i++)
        {
            cin >> t;
            A[t-1][i] = 1;
        }
        getline(cin,key);
        getline(cin,key);
        arrayPow(A,m);

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(res[i][j])
                {
                    cout << key[j];
                    break;
                }
            }
        }
        cout << endl;
    }
    return 0;
}