2020 BIT冬训-二分三分快速幂矩阵 L - Decode the Strings HDU - 2371

标签：ini   repr   循环   algorithm   scan   value   size   desc   color   

Problem Description

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,...,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.
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);
    }
}

2020 BIT冬训-二分三分快速幂矩阵 L - Decode the Strings HDU - 2371

标签：ini   repr   循环   algorithm   scan   value   size   desc   color   

原文地址：https://www.cnblogs.com/mikku39/p/14409734.html