Power Tower（广义欧拉降幂）
作者：互联网
题意：https://codeforc.es/contest/906/problem/D
计算区间的：
ai ^ ai+1 ^ ai+2.......ar 。
思路：
广义欧拉降幂：
注意是自下而上递归使用欧拉降幂，比如求：a^b^c == a^(b^c%phi(mod)+?) == a^(b^(c%phi(phi(mod))+?+?)
而不是：a^b^c == a^b^(c%phi(mod)+?) == a^(b^(c%phi(mod)+?)%phi(mod)+?) 这样本身就是不对的，次方不是这么算的。
注意：因为判断要不要+phi（mod），所有快速幂里面就要开始搞搞，自己标个flag，或者直接重定义Mod == return x>=m?x%m+m:x;
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
2 #include <cstdio>//sprintf islower isupper
3 #include <cstdlib>//malloc exit strcat itoa system("cls")
4 #include <iostream>//pair
5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\草稿.txt","r",stdin);
6 #include <bitset>
7 #include <map>
8 //#include<unordered_map>
9 #include <vector>
10 #include <stack>
11 #include <set>
12 #include <string.h>//strstr substr
13 #include <string>
14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
15 #include <cmath>
16 #include <deque>
17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
18 #include <vector>//emplace_back
19 //#include <math.h>
20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
23 #define fo(a,b,c) for(register int a=b;a<=c;++a)
24 #define fr(a,b,c) for(register int a=b;a>=c;--a)
25 #define mem(a,b) memset(a,b,sizeof(a))
26 #define pr printf
27 #define sc scanf
28 #define ls rt<<1
29 #define rs rt<<1|1
30 typedef long long ll;
31 void swapp(int &a,int &b);
32 double fabss(double a);
33 int maxx(int a,int b);
34 int minn(int a,int b);
35 int Del_bit_1(int n);
36 int lowbit(int n);
37 int abss(int a);
38 //const long long INF=(1LL<<60);
39 const double E=2.718281828;
40 const double PI=acos(-1.0);
41 const int inf=(1<<30);
42 const double ESP=1e-9;
43 //const int mod=(int)1e9+7;
44 const int N=(int)1e6+10;
45
46 ll a[N];
47 map<ll,ll>mp;
48 long long phi(long long n)//a^(b mod phi(c)+phi(c)) mod c
49 {
50 if(mp.count(n))return mp[n];//记忆化;
51 long long i,rea=n,temp=n;
52 for(i=2;i*i<=n;i++)
53 {
54 if(n%i==0)
55 {
56 rea=rea-rea/i;
57 while(n%i==0)
58 n/=i;
59 }
60 }
61 if(n>1)
62 rea=rea-rea/n;
63 mp[temp]=rea;
64 return rea;
65 }
66 ll Mod(ll x, ll m)
67 {
68 return x>=m?x%m+m:x;
69 }
70 long long qpow(long long a,long long b,long long mod)
71 {
72 long long ans,fl=0;
73 // ll ta=a,tb=b,tta=a;
74 ans=1;
75 while(b!=0)
76 {
77 if(b&1)
78 {
79 if(ans*a>=mod)fl=1;
80 ans=ans*a%mod;
81 }
82 b/=2;
83 if(a*a>=mod)fl=1;
84 a=a*a%mod;
85 }
86 return ans+(fl?mod:0);
87 }
88 ll solve(int l,int r,ll mod)//返回l~r计算结果; 听说phi(phi(mod))~==~mod/2所有最多log次;
89 {
90 if(l==r||mod==1)return Mod(a[l],mod);//任何数%1都是0,不用再算了;
91 return qpow(a[l],solve(l+1,r,phi(mod)),mod);//假设我已经知道了l+1~r的结果:递归下去;
92 }
93
94 int main()
95 {
96 int n;
97 ll p;
98 sc("%d%lld",&n,&p);
99 fo(i,1,n)sc("%lld",&a[i]);
100 int ask;sc("%d",&ask);
101 while(ask--)
102 {
103 int l,r;
104 sc("%d%d",&l,&r);
105 pr("%lld\n",solve(l,r,p)%p);
106 }
107 return 0;
108 }
109
110 /**************************************************************************************/
111
112 int maxx(int a,int b)
113 {
114 return a>b?a:b;
115 }
116
117 void swapp(int &a,int &b)
118 {
119 a^=b^=a^=b;
120 }
121
122 int lowbit(int n)
123 {
124 return n&(-n);
125 }
126
127 int Del_bit_1(int n)
128 {
129 return n&(n-1);
130 }
131
132 int abss(int a)
133 {
134 return a>0?a:-a;
135 }
136
137 double fabss(double a)
138 {
139 return a>0?a:-a;
140 }
141
142 int minn(int a,int b)
143 {
144 return a<b?a:b;
145 }
标签：phi,return,Power,降幂,int,long,Tower,include,mod 来源： https://www.cnblogs.com/--HPY-7m/p/11444923.html