逆置换 poj 1721

题意：

给n张已经被洗好的卡片，他们被洗过s次，每次洗卡片操作是把a[i]换成a[a[i]],即一个置换的平方，求出卡片被放入洗牌机之前是什么样的。

可以把置换循环的长度求出来，设为a，原始置换置换s次后得到当前置换，那么当前置换再置换a-s次记得到了原始置换。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/
int c[1005];

int n,s;
int f[1005];
int b[1005];
int lenth()
{int cnt=0;
while(1)
{REP(i,1,n+1)
c[i]=b[b[i]];
cnt++;
int ok=1;
REP(i,1,n+1)
if(c[i]!=f[i])
    {
        ok=0;
        break;
    }
if(ok)
    break;
REP(i,1,n+1)
b[i]=c[i];

}
return cnt;
}
int main()
{RII(n,s);
REP(i,1,n+1)
{
    RI(f[i]);
    c[i]=f[i];
    b[i]=f[i];
}
int a=lenth();
int left=a-s%a;
REP(i,1,n+1)
b[i]=f[i];
while(left--)
{
    REP(i,1,n+1)
    f[i]=b[b[i]];
    REP(i,1,n+1)
    b[i]=f[i];
}
REP(i,1,n+1)
printf("%d\n",f[i]);
    return 0;
}