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): 669    Accepted Submission(s): 121

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

BestCoder Round #51 (div.2)

 

先求出每个环的大小Ni,然后求所有Ni的最小公倍数。

那么就是对于每个质因数，求出现在Ni的最多的次数，然后做乘法即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
#define ll unsigned long long
#define maxn 3000007
int check[maxn];
vector<int> pri;
void init(){
    memset(check,0,sizeof(check));
    pri.clear();
    for(int i = 2;i < maxn; i++){
        if(check[i] == 0) {
            pri.push_back(i);
        }
        for(int j = 0;j < pri.size() ; j++){
            if(i*pri[j] >= maxn) break;
            check[i*pri[j]] = 1;
            if(i%pri[j] == 0) break;
        }
    }
}
int getn(){
    int a = 0;
    char x;
    do{
        x = getchar();
        if(x == ' ' || x == '\n') return a;
        a = a*10+x-'0';
    }while(1);
    return 1;
}

int num[maxn];
int ok[maxn];

vector<int> hehe;

ll mod = 3221225473ll;
int pnum[maxn];
ll cal(int u,int p){
    ll ans = 1;
    ll v = u;
    while(p){
        if(p & 1) ans = ans*v % mod;
        p/=2;
        v = v*v%mod;
    }
    return ans;
}


int main(){
    init();
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        getchar();
        for(int i = 1;i <= n; i++)
            num[i] = getn();
        for(int i = 1;i <= n; i++)
            ok[i] = pnum[i] = 0;
        hehe.clear();
        int mu = 1;
        for(int i = 1;i <= n; i++){
            if(ok[i] != 1){
                int u = i;
                int x = 0;
                while(ok[u] == 0){
                    x++;
                    ok[u] = 1;
                    u = num[u];
                }
                hehe.push_back(x);
            }
        }

        for(int i = 0;i < hehe.size(); i++){
            for(int j = 0;j < pri.size() && pri[j] <= hehe[i]; j++){
                if(hehe[i] % pri[j] == 0){
                    int x = 0;
                    while(hehe[i] % pri[j] == 0){
                        hehe[i] /= pri[j];
                        x++;
                    }
                    pnum[pri[j]] = max(pnum[pri[j]],x);
                    mu = max(mu,pri[j]);
                }
            }
        }

        ll ans = 1;
        for(int i = 1;i <= mu; i++){
            if(pnum[i] != 0)
            ans = ans*cal(i,pnum[i])%mod;
        }


        printf("%I64d\n",ans);

    }
    return 0;
}