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  • super_log (广义欧拉降幂)(2019南京网络赛)

    题目:

    In Complexity theory, some functions are nearly O(1)O(1), but it is greater then O(1)O(1). For example, the complexity of a typical disjoint set is O(nα(n))O(nα(n)). Here α(n)α(n) is Inverse Ackermann Function, which growth speed is very slow. So in practical application, we often assume α(n) le 4α(n)4.

    However O(α(n))O(α(n)) is greater than O(1)O(1), that means if nn is large enough, α(n)α(n) can greater than any constant value.

    Now your task is let another slowly function log*log∗ xx reach a constant value bb. Here log*log∗ is iterated logarithm function, it means “the number of times the logarithm function iteratively applied on xx before the result is less than logarithm base aa”.

    Formally, consider a iterated logarithm function log_{a}^*loga

    Find the minimum positive integer argument xx, let 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(Tle 300)T(T300) indicating the number of test cases.

    Each of the following lines contains 33 integers aa, bb and mm.

    1 le a le 10000001a1000000

    0 le b le 10000000b1000000

    1 le m le 10000001m1000000

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

    Output

    For each test case, output xx mod mm in a single line.

    Hint

    In the 4-th4th query, a=3a=3 and b=2b=2. Then 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)=2b, so the output is 2727 mod 16 = 1116=11.

    样例输入

    5
    2 0 3
    3 1 2
    3 1 100
    3 2 16
    5 3 233

    样例输出

    1
    1
    3
    11
    223

    解题报告:严重怀疑这场比赛的出题方,读题实在是太困难了,看了半个小时才理解了要求解什么东西,一看就是一个广义欧拉降幂,然后就直接去写的之前的模板,求解无穷个2的幂次,
    但是修修改改的过了样例,但是忘记了广义欧拉降幂的实现,疯狂wa,在比赛结束后找了好久才发现是没有去使用广义欧拉降幂的性质,忘记在该加欧拉(m)的地方进行加了,隔壁队的
    数论手,给了一个能够实现这个功能的快速幂,稍微一修改就A了。这个需要掌握了,当时时间太紧,头脑已经混了。

    ac代码:


     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
    ",1%m);
    93             continue;
    94         }
    95         ans=0;
    96         printf("%lld
    ",slove(m,1)%m);
    97     }
    98 }


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  • 原文地址:https://www.cnblogs.com/Spring-Onion/p/11444096.html
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