【置换群】 HDOJ 1439 && POJ 1026 Cipher

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;
}


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