求出置换群的每一个子群的循环节,然后对每一个置换模拟一遍即可。。。
#include <iostream> #include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits> #include <cstdlib> #include <cmath> #include <time.h> #define maxn 100005 #define maxm 40005 #define eps 1e-10 #define mod 1000000007 #define INF 999999999 #define lowbit(x) (x&(-x)) #define mp mark_pair #define ls o<<1 #define rs o<<1 | 1 #define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R typedef long long LL; //typedef int LL; using namespace std; LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;} void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();} LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);} // head int change[maxn], next[maxn], vis[maxn]; int c[maxn], f[maxn]; char s[maxn], t[maxn]; int n, m, k; void work(void) { int now, cnt, len; memset(vis, 0, sizeof vis); for(int i = 1; i <= n; i++) if(!vis[i]) { now = i, cnt = 1, f[now] = i; while(change[now] != i) vis[now] = 1, now = change[now], cnt++, f[now] = i; vis[now] = 1, c[i] = cnt; } while(scanf("%d", &k), k != 0) { for(int i = 1; i <= n; i++) next[i] = i; for(int i = 1; i <= n; i++) { int t = k % c[f[i]]; while(t--) next[i] = change[next[i]]; } getchar(), gets(s+1); len = strlen(s+1); for(int i = len+1; i <= n; i++) s[i] = ' '; for(int i = 1; i <= n; i++) t[next[i]] = s[i]; t[n+1] = '\0'; puts(t+1); } } int main(void) { while(scanf("%d", &n), n != 0) { for(int i = 1; i <= n; i++) scanf("%d", &change[i]); work(); puts(""); } return 0; }