super_log (Generalized Euler descending) (2019 Nanjing network game)

topic:

In Complexity theory, some functions are nearly $\le 4α(n)≤4.$

However

Now your task is let another slowly function

Formally, consider a iterated logarithm function $^*loga∗$

Find the minimum positive integer argument $log_{a}^* (x) \ge bloga∗(x)≥b. The answer may be very large, so just print the result xx after mod mm.$

Input

The first line of the input is a single integer $T(T\le 300)T(T≤300) indicating the number of test cases.$

Each of the following lines contains

$\ has the \ the 1000000 1 ≤ a ≤ 1 0 0 0 0 0 0$

$\ b \ the 1000000 0 ≤ b ≤ 1 0 0 0 0 0 0$

$\ m \ the 1000000 1 ≤ m ≤ 1 0 0 0 0 0 0$

Note that if a==1, we consider the minimum number x is 1.

Output

For each test case, output

Hint

In the $log_{3}^* (27) = 1+ log_{3}^* (3) = 2 + log_{3}^* (1)=3+(-1)=2 \ge blog3∗(27)=1+log3∗(3)=2+log3∗(1)=3+(−1)=2≥b, so the output is 2727 mod 16 = 1116=11.$

Sample input

Sample Output

1 
1 
3 
11 
223 

problem-solving report: serious doubts about the topic and square the match, the question is too difficult to read, and read half an hour before to understand what the requirements of the solution, a look that is a generalized Euler descending, then the template prior written directly to solving an infinite number of powers of 2, 
but minor changes over the sample, but forget the generalized Euler descending implementation, crazy wa, after the game for a long time only is found not to use the properties of the generalized Euler descending, and forget to add in the Jiaou La (m) where the next team of 
number theory hand, gave a quick power can achieve this functionality, a slight modification a a. The need to know, at that time the tight schedule, the mind has been mixed. 

ac Code:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<map>
 7 #include<vector>
 8 using namespace std;
 9 typedef long long ll;
10 
11 const int maxn=1e7+10;
12 vector<int> prime;
13 int phi[maxn];
14 
15 void db_Onion()
16 {
17     phi[1]=1;
18     for(int i=2;i<maxn;i++)
19     {
20         if(phi[i]==0)
21         {
22             prime.push_back(i);
23             phi[i]=i-1;
24         }
25         for(int j=0;j<prime.size()&&prime[j]*i<maxn;j++)
26         {
27             if(i%prime[j]==0)
28             {
29                 phi[i*prime[j]]=phi[i]*prime[j];
30                 break;
31             }
32             else
33             {
34                 phi[i*prime[j]]=phi[i]*(prime[j]-1);
35             }
36         }
37     }
38 }
39 ll a,b,m;
40 ll cal(ll x,ll m)
41 {
42     if(x<m)
43         return x;
44     else
45         return x%m+m;
46 }
47 
48 ll ans;
49 
50 ll ksm(ll r,ll n,ll m)
51 {
52     ll d=r,ans=1;
53     bool f=0;
54     if(r==0&&n==0)return 1;
55     while(n){
56         if(n&1){
57             ans=ans*d;
58             if(ans>=m){
59                 ans%=m;f=1;
60             }
61         }
62         d=d*d;
63         if(n>1&&d>=m){
64             d%=m;f=1;
65         }
66         n>>=1;
67     }
68     if(f)ans+=m;
69     return ans;
70 }
71 ll slove(ll p,int tot)
72 {
73     if(tot==b+1)
74         return 1;
75     if(p==1)
76         return 1;
77     
78     ll tmp=slove(phi[p],tot+1); 
79     return ksm(a,tmp,p);        
80 }
81 
82 int main()
83 {
84     db_Onion();
85     int t;
86     scanf("%d",&t);
87     while(t--)
88     {
89         scanf("%lld%lld%lld",&a,&b,&m);
90         if(b==0||a==1)
91         {
92             printf("%lld\n",1%m);
93             continue;
94         }
95         ans=0;
96         printf("%lld\n",slove(m,1)%m);
97     }
98 }