HDU-6189

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题意

给出 $n(\le 30), a(1\le a\le 1e9)$ , 求在 $[1,2^n]$ 范围内有多少个 $b$ 能够满足 $a^{b} \equiv b^{a}(\bmod 2^n)$

题解

https://blog.csdn.net/hyesuixin/article/details/77855821

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define memcp(a,b) memcpy(a,b,sizeof(b))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pu_b push_back
#define pu_f push_front
#define po_b pop_back
#define po_f pop_front
#define fi first
#define se second
#define whiel while
#define retrun return
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
void out(int a){printf("%d ",a);}
void outln(int a){printf("%d\n",a);}
void out(ll a){printf("%lld ",a);}
void outln(ll a){printf("%lld\n",a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 2.1e5;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps;}
int db_cmp(db a, db b){ return sign(a-b); }

ll qPow(ll a, ll b, ll c) {  //求(a^b) % c
    ll ret = 1;
    while (b) {
        if (b & 0x1) ret = ret * a % c;
        a = a * a % c;
        b >>= 1;
    }
    return ret;
}
int main() {
    int n,a;
    whiel(~in(n,a)){
        int ta=a,k=0,maxn=1<<n;
        whiel(ta%2==0){
            k++;
            ta/=2;
        }
        if(k&&n){
            itn kk=n/a;
            if(n%a)kk++;
            int ans=0;
            int low=n/k;
            if(n%k==0)low--;
            for1(i,low){
                if(qPow(a,i,maxn)==qPow(i,a,maxn))ans++;
            }
            kk=1<<kk;
            ans+=(1<<n)/kk-low/kk;
            outln(ans);
        }else { 
            puts("1");
        }
    }
    return 0;
}