HDU 5392 Infoplane in Tina Town

Infoplane in Tina Town

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 805    Accepted Submission(s): 168

Problem Description

There is a big stone with smooth surface in Tina Town. When people go towards it, the stone surface will be lighted and show its usage. This stone was a legacy and also the center of Tina Town’s calculation and control system. also,
it can display events in Tina Town and contents that pedestrians are interested in, and it can be used as public computer. It makes people’s life more convenient (especially for who forget to take a device).



Tina and Town were playing a game on this stone. First, a permutation of numbers from
1
to n
were displayed on the stone. Town exchanged some numbers randomly and Town recorded this process by macros. Town asked Tine,”Do you know how many times it need to turn these numbers into the original permutation by executing this macro? Tina didn’t know the
answer so she asked you to find out the answer for her.



Since the answer may be very large, you only need to output the answer modulo
3∗230+1=3221225473
(a prime).

 

Input

The first line is an integer
T
representing the number of test cases. T≤5



For each test case, the first line is an integer n
representing the length of permutation. n≤3∗106



The second line contains n
integers representing a permutation A1...An.
It is guaranteed that numbers are different each other and all
Ai
satisfies ( 1≤Ai≤n
).

 

Output

For each test case, print a number
ans
representing the answer.

 

Sample Input

2
3
1 3 2
6
2 3 4 5 6 1

 

Sample Output

2
6

 

Source

field=problem&key=BestCoder+Round+%2351+%28div.2%29&source=1&searchmode=source">BestCoder Round #51 (div.2)

 

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给出一个序列，求变换几次能够回到原来的位置。比方 1 3 2 ，3 不在原来的位置，变到3位置。次数加1，2变到2，次数+1.得到2.。

做法就是分解循环长度。然后求下最小公倍数。

可是不能直接用lcm求最小公倍数。

。我们能够考虑用质数分解来求，即公共的质因子乘每一个数本身的质因子。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
int a[3000010];
int vis[3000010];
int b[3000010];
const ll mod=3221225473;
inline int read()
{
    char ch;
    for (ch=getchar(); ch<48||ch>57;) ch=getchar();
    int d=0;
    for (; ch>47&&ch<58; ch=getchar()) d=d*10+ch-48;
    return d;
}
ll gcd(ll a,ll b)
{
    if(b==0)
        return a;
    else
        return gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    int t,n,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1; i<=n; i++)
            a[i]=read();
        memset(vis,0,sizeof(vis));
        memset(b,0,sizeof(b));
        for(i=1; i<=n; i++)
        {
            if(!vis[i])
            {

                int t=i;
                int s=0;
                while(!vis[t])
                {
                    s++;
                    vis[t]=1;
                    t=a[t];
                }
                for(j=2; j*j<=s; j++)
                {
                    int cnt=0;
                    while(s%j==0)
                    {
                         cnt++;
                        s/=j;
                    }
                    b[j]=max(b[j],cnt);  //统计公共的质因子。

}
                if(s>1)
                    b[s]=max(b[s],1);
            }
        }
        //  cout<<lcm(121,11)<<endl;
        ull ans=1;
        for(i=2; i<=n; i++)
            for(j=1; j<=b[i]; j++)
                    ans=ull(ans)*i%mod;
        cout<<ans<<endl;
    }
}