bzoj 3643Phi的反函数

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<queue>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<map>
 9 #define N 50005
10 using namespace std;
11 long long sq,n,ss[N],zz,ans=4294967296;
12 bool fss[N];
13 bool pri(long long x)
14 {
15     for(int i=2;i<=sqrt(x);i++)
16         if(x%i==0)return 0;
17     return 1;
18 }
19 void dfs(long long sum,long long wz,long long now)
20 {
21     if(sum>ans)return;
22     if(now==1)
23     {
24         ans=min(ans,sum);
25         return;
26     }
27     if(now>sq&&pri(now+1))ans=min(ans,(now+1)*sum);
28     for(int i=wz;i<=zz&&ss[i]-1<=sq&&ss[i]-1<=now;i++)
29     {
30         if(now%(ss[i]-1)==0)
31         {
32             int x=now/(ss[i]-1),y=sum*ss[i];
33             dfs(y,i+1,x);
34             while(x%ss[i]==0)
35             {
36                 x/=ss[i],y*=ss[i];
37                 dfs(y,i+1,x);
38             }
39         }
40     }
41 }
42 int main()
43 {
44     scanf("%lld",&n);
45     sq=sqrt(n);
46     for(int i=2;i<=sqrt(n);i++)
47     {
48         if(!fss[i])
49         {
50             zz++;
51             ss[zz]=i;
52         }
53         for(int j=1;j<=zz&&i*ss[j]<N;j++)
54         {
55             fss[i*ss[j]]=1;
56             if(i%ss[j]==0)break;
57         }
58     }
59     dfs(1,1,n);
60     if(ans<2147483648ll) 
61         printf("%lld\n",ans);
62     else printf("-1\n");
63     return 0;
64 }