hdu5392

求循環節加上線性求lcm 即lcm等於公共的質因子乘每個數本身的質因子。

代碼：

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <map>
#include <set>
#define sss(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a) memset(a,0,sizeof(a))
#define ss(a,b) scanf("%d%d",&a,&b)
#define s(a) scanf("%d",&a)
#define p(a) printf("%d\n", a)
#define INF 0x3f3f3f3f
#define w(a) while(a)
#define PI acos(-1.0)
#define LL long long
#define eps 10E-9
#define N 3000010
#define mod 3221225473
const int SIGMA_SIZE=26;
const int MAXN=100010;
const int MAXNODE=600010;
using namespace std;
void mys(int& res) {
    int flag=0;
    char ch;
    while(!(((ch=getchar())>='0'&&ch<='9')||ch=='-'))
        if(ch==EOF)  res=INF;
    if(ch=='-')  flag=1;
    else if(ch>='0'&&ch<='9')  res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+ch-'0';
    res=flag?-res:res;
}
void myp(int a) {
    if(a>9)
        myp(a/10);
    putchar(a%10+'0');
}
/*************************THE END OF TEMPLATE************************/
int arr[N];
int lcm[N];
bool vis[N];
int main() {
    int n,t;
    s(t);
    w(t--) {
        s(n);
        mem(vis);
        bool flag = false;
        for(int i=1; i<=n; i++) {
            s(arr[i]);
            if(i>1 && arr[i]-arr[i-1] != 1) flag = true;
        }
        mem(lcm);
        if(flag) {
            for(int i=1; i<=n; i++) {
                if(!vis[i]) {
                   int a=i;
                   int cnt=0;
                   w(!vis[a]){
                       cnt++;
                       vis[a]=1;
                       a=arr[a];
                   }
                   for(int j=2; j*j<=cnt; j++){
                        int ans=0;
                        w(cnt%j==0){
                            ans++;
                            cnt/=j;
                        }
                        lcm[j]=max(lcm[j],ans);
                   }
                  if(cnt>1) lcm[cnt] = max(lcm[cnt], 1);
                }
            }
           unsigned LL sum=1;
           for(int i=2; i<=n; i++){
                for(int j=1; j<=lcm[i]; j++) sum =(unsigned LL)(sum)*i%mod;
          }
          printf("%u\n",sum);
        } else p(0);
    }
    return 0;
}