传送门：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3255
题解：LRJ训练之南p147
说一下碎玉置换A=(a1 a2 a3)(b1 b2 b3 b4)的平方怎么得出来的：
(a1 a2 a3)(a1 a2 a3)=>a1->a2->a3,a2->a3->a1,a3->a1->a2=(a1 a3 a2)
(b1 b2 b3 b4)(b1 b2 b3 b4)=> b1->b2->b3,b2->b3->b4,b3->b4->b1,b4->b1->b2=(b1 b3)(b2 b4)
#include <iostream> #include <cstdio> #include <cstring> using namespace std; char s[30]; int vis[30],a[30]; int main() { int T; scanf("%d",&T); while(T--) { scanf("%s",s); memset(vis,0,sizeof(vis)); memset(a,0,sizeof(a)); for(int i=0;i<26;i++) { if(!vis[i]) { int j=i,n=0; do{ vis[j]=1; j=s[j]-'A'; n++; }while(j!=i); a[n]++; } } int ok=1; for(int i=2;i<=26;i+=2) if(a[i]%2==1)ok=0; if(!ok)printf("No\n"); else printf("Yes\n"); } return 0; }