UVALA 3641 Leonardo's Notebook

Polya定理应用:

题意：给出一个置换B，问有没有一个置换A，使得A^2=B。

思路：对于置换的循环节，比如(a1,a2,a3)和(b1,b2,b3,b4)，那么对于奇数长度的循环节，发现进行两次乘法以后还是奇数长度，偶数长度的循环节分解为两个长度相同的循环节。那么就可以对B中的循环节进行判断，奇数长度的不同管，这个总能找到满足要求的，偶数循环节的数量也要是偶数，这样才能两两配对。

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3641 Leonardo’s Notebook

— I just bought Leonardo’s secret notebook! Rare object collector
Stan Ucker was really agitated but his friend, special
investigator Sarah Keptic was unimpressed.
— How do you know it is genuine?
— Oh, it must be, at that price. And it is written in the da
Vinci code. Sarah browsed a few of the pages. It was obvious
to her that the code was a substitution cipher, where each
letter of the alphabet had been substituted by another letter.
— Leonardo would have written the plain-text and left it to
his assistant to encrypt, she said. And he must have supplied
the substitution alphabet to be used. If we are lucky, we can
find it on the back cover!
She turned up the last page and, lo and behold, there was
a single line of all 26 letters of the alphabet:
QW ERT Y UIOP ASDF GHJKLZXCV BNM
— This may be Leonardo’s instructions meaning that each A in the plain-text was to be replaced
by Q, each B with W, etcetera. Let us see…
To their disappointment, they soon saw that this could not be the substitution that was used in the
book. Suddenly, Stan brightened.
— Maybe Leonardo really wrote the substitution alphabet on the last page, and by mistake his
assistant coded that line as he had coded the rest of the book. So the line we have here is the result of
applying some permutation TWICE to the ordinary alphabet!
Sarah took out her laptop computer and coded fiercely for a few minutes. Then she turned to Stan
with a sympathetic expression.
— No, that couldn’t be it. I am afraid that you have been duped again, my friend. In all probability,
the book is a fake.
Write a program that takes a permutation of the English alphabet as input and decides if it may
be the result of performing some permutation twice.
Input
The input begins with a positive number on a line of its own telling the number of test cases (at most
500). Then for each test case there is one line containing a permutation of the 26 capital letters of the
English alphabet.
Output
For each test case, output one line containing ‘Yes’ if the given permutation can result from applying
some permutation twice on the original alphabet string ABC…XYZ, otherwise output ‘No’.
ACM-ICPC Live Archive: 3641 – Leonardo’s Notebook 2/2
Sample Input
2
QWERTYUIOPASDFGHJKLZXCVBNM
ABCDEFGHIJKLMNOPQRSTUVWXYZ
Sample Output
No
Yes

/* ***********************************************Author        :CKbossCreated Time  :2015年05月06日 星期三 21时01分41秒File Name     :UVALA3641.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;char str[30];bool vis[30];int len[50];int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        scanf("%s",str);        memset(vis,0,sizeof(vis));        memset(len,0,sizeof(len));        for(int i=0;i<26;i++)        {            int id=str[i]-'A';            if(vis[id]==true) continue;            int loop=0;            do            {                loop++;                vis[id]=true;                id=str[id]-'A';            }while(vis[id]==false);            len[loop]++;        }        bool flag=true;        for(int i=2;i<=40&&flag;i+=2)        {            if(len[i]%2!=0) flag=false;         }        if(flag) puts("Yes");        else puts("No");    }    return 0;}