poj 2369 Permutations

Permutations

题意:输入长度为n(1 <= n <= 1000)的置换，定义P(n) = P1(n), Pk(n) = P(Pk-1(n));问K为多少时，置换之后变成f[i] = i的置换；

水题:用的就是循环相乘，元素对应关系的理解；即(a1,a2,a3...,an)经过一次相乘之后，a1->a3(a1->a2->a3);所以对于每个循环，要使得a1->a1,该循环K的值就是循环的大小；所以直接求出所有循环大小的lcm即可；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
typedef __int64 ll;
template<typename T>
void read1(T &m)
{
    T x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+‘0‘);
}
int a[1010];
int __gcd(int a,int b)
{
    return b == 0?a:__gcd(b,a%b);
}
int lcm(int a,int b)
{
    return a/__gcd(a,b)*b;
}
int vis[1010],B[1010];
int main()
{
    int N;
    read1(N);
    rep1(i,1,N) read1(B[i]);
    int ans = 1;
    rep1(i,1,N){
        int tmp = i,cnt = 0;
        do{
            vis[tmp] = ++cnt;
            tmp = B[tmp];
        }while(!vis[tmp]);
        ans = lcm(ans,cnt);
    }
    printf("%d\n",ans);
    return 0;
}