2020 BIT冬训-二分三分快速幂矩阵 L - Decode the Strings HDU - 2371
时间：2021-02-19 13:18:41
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Problem Description
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,...,xn be the sequence of characters of the string to be encoded.
1. Choose an integer m and n pairwise distinct numbers p1,p2,...,pn 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 yi to xpi, and then for 1 ≤ i ≤ n replace xi by yi.
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, ..., pn 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?
Let x1,x2,...,xn be the sequence of characters of the string to be encoded.
1. Choose an integer m and n pairwise distinct numbers p1,p2,...,pn 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 yi to xpi, and then for 1 ≤ i ≤ n replace xi by yi.
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, ..., pn 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?
InputThe 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,...,pn (1 ≤ pi ≤ 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.
OutputFor 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?
这题是寻找循环节。（跟他所在的题单好像完全没关系啊）
循环节就是重复一定次数后达到一种循环的状态。由于字符串的转换关系是不变的。固必然会出现循环。（具体论证懒得算了。鸽巢定理（吧））
寻找字符串的循环节比较困难且复杂且耗时。由于每个字符串最多只有80个字符。故我们可以对字符串内每一个字符寻找其循环节即可。
AC代码如下：
#include<cstdio> #include<algorithm> #include<cstring> using namespace std; int n,m,p[85],temp,pos[85][85],cnt; char s[85],ans[85]; int main(){ while(scanf("%d%d",&n,&m),n){ for(int i=0;i<n;i++){ scanf("%d",&p[i]); p[i]--; } getchar(); gets(s); for(int i=0;i<n;i++){ cnt=0; temp=p[i]; pos[i][cnt++]=i; while(temp!=i){ pos[i][cnt++]=temp;//记录i在第cnt次循环在 什么位置 temp=p[temp]; } ans[pos[i][m%cnt]]=s[i]; } ans[n]=‘\0‘; printf("%s\n",ans); } }
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