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 0x3f3f3f3fusing 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);    }}