2017 多校 水题

首先预处理每个数的因数有哪些，然后分别记录a和b中有几个环。 最后就是两层循环，注意取膜。
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<string.h>
#include<queue>
#include<stack>
#include<map>
#define mod 1000000007
#define INF 0x3f3f3f3f
using namespace std;
int n,m;
const int N = 100000+5;
struct node
{
    int num,fa;
}a[N],b[N];
struct po
{
    vector<int>v;
}po[N];
int va[N],vb[N];
int cira[N],cb[N];
void init()
{
    for(int i = 1; i <= 50000; i++)
    {
        for(int j = i; j <= 100000; j+= i)
        {
              po[j].v.push_back(i);
        }
    }
    return ;
}
int main()
{
    init();
    int tim = 1;
    while(~scanf("%d%d",&n,&m))
    {
         for(int i = 0; i < n; i++)
            scanf("%d",&a[i].num);
         for(int i = 0; i < m; i++)
            scanf("%d",&b[i].num);
         memset(va,0,sizeof(va));
         memset(vb,0,sizeof(vb));
         for(int i = 0; i < n; i++)
         {
             int te = a[i].num;
             a[te].fa = i;
         }
         for(int i = 0; i < m; i++)
         {
             int te = b[i].num;
             b[te].fa = i;
         }
         memset(cb,0,sizeof(cb));
        memset(cira,0,sizeof(cira));
     int cnt = 0;
     for(int i = 0; i < n; i++)
         {
             if(!va[i])
             {
                 va[i] = 1;
                 int ff = a[i].fa;
                 int root = i;
                  int sum = 1;
                 while(ff != root)
                 {
                     va[ff] = 1;
                     ff = a[ff].fa;
                     sum ++;
                 }
                 cira[cnt++] = sum;
             }
         }
         for(int i = 0; i < m; i++)
         {
             if(!vb[i])
             {
                 vb[i] = 1;
                 int ff = b[i].fa;
                 int root = i;
                 int sum = 1;
                 while(ff != root)
                 {
                     vb[ff] = 1;
                     ff = b[ff].fa;

                     sum++;
                 }
              //   cout << sum << endl;
                 cb[sum] += 1;
             }
         }
         long long ans = 1;
         for(int i = 0; i < cnt; i++)
         {
             int nn = cira[i];
             long long tem = 0;
             for(int i = 0; i < po[nn].v.size(); i++)
             {
                 int j = po[nn].v[i];
                 tem += cb[j]*j;
            //     cout << j <<"        " << tem << endl;
             }
             ans = (ans%mod*tem%mod)%mod;
         }
         printf("Case #%d: %lld\n",tim++,ans);
    }
}