Ultraviolet A 12103-Leonardo's notebook (replacement)

Link to the question: Ultraviolet A 12103-Leonardo's notebook

Given a replacement of 26 letters, ask if there is a replacement for a, so that A2 = B

Solution: break down a given replacement into several irrelevant cycles. When the length n of a loop is an odd number, it can be obtained by the product of two cycles whose length is N, it can also be split from two cycles with a length of 2n. If n is an even number, it can only be split from two cycles with a length of 2n, therefore, you can check whether there are an odd number with an even cycle length.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 30;int main () {    char s[maxn];    int cas, v[maxn], c[maxn];    scanf("%d", &cas);    while (cas--) {        memset(v, 0, sizeof(v));        memset(c, 0, sizeof(c));        scanf("%s", s);        for (int i = 0; i < 26; i++) {            if (v[i])                continue;            int j = i, ret = 0;            do {                v[j] = 1;                j = s[j] - ‘A‘;                ret++;            } while (i != j);            c[ret]++;        }        bool flag = true;        for (int i = 2; i < 26; i += 2)            if (c[i]&1)                flag = false;        printf("%s\n", flag ? "Yes" : "No");    }    return 0;}