歐拉降冪及其擴展歐拉降冪

歐拉降冪：

從公式來看，需要使用快速冪運算和歐拉函數

#include<bits/stdc++.h>

using namespace std;
typedef __int64 LL;

const int maxn = 1e6+100;

char b[maxn];

//歐拉函數
LL ouler(LL n){
    LL ans = n, a = n;
    for(LL i = 2; i * i <= a; i++){
        if(a % i == 0){
            ans -= ans / i;
            while(a % i == 0){
                a /= i;
            }
        }
    }
    if(a > 1){
        ans -= ans / a;
    }
    return ans;
}

//快速冪
LL quick_mod(LL a, LL b, LL mod){
    LL res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

LL solve(LL a, char b[], LL mod){
    int len = strlen(b);
    LL phi = ouler(mod);

    LL tmp = 0;
    for(int i = 0; i < len; i++){
        tmp = tmp * 10 + (b[i] - '0');
        if(tmp > mod) break;
    }

    if(tmp <= mod) return quick_mod(a, tmp, mod);
    else{
        tmp = 0;
        for(int i = 0; i < len; i++)
            tmp = ( tmp * 10 + (b[i] - '0') ) % phi;
        return quick_mod(a, tmp + phi, mod);
    }
}

//  A ^ B % p
int main()
{
    LL a, c;
    LL n = 0, tmp;
    while(scanf("%lld", &a) != EOF){
        scanf("%s%lld", b, &c);
        printf("%lld\n", solve(a, b, c));
    }
    return 0;
}

 

擴展歐拉降冪

求   b個a的次冪

解法：

對於，我們可以直接使用歐拉降冪去求，但這個式子是a次冪，又該怎麼辦？

那麼他的下一步一定還是遞歸的形式已知往下，直到q == 1 時，向上返回結果，所以只需在歐拉降冪的基礎上加一個dfs

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;

const int maxn = 1e6+100;

char b[maxn];

//歐拉函數
LL ouler(LL n){
    LL ans = n, a = n;
    for(LL i = 2; i * i <= a; i++){
        if(a % i == 0){
            ans -= ans / i;
            while(a % i == 0){
                a /= i;
            }
        }
    }
    if(a > 1){
        ans -= ans / a;
    }
    return ans;
}

//快速冪
LL quick_mod(LL a, LL b, LL mod){
    LL res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

LL solve2(LL a, LL b, LL mod){
    LL phi = ouler(mod);
    if(b <= mod) return quick_mod(a, b, mod);
    else{
        a = a % phi;
        return quick_mod(a, b+phi, mod);
    }
}

LL solve(LL a, char b[], LL mod){
    int len = strlen(b);
    LL phi = ouler(mod);

    LL tmp = 0;
    for(int i = 0; i < len; i++){
        tmp = tmp * 10 + (b[i] - '0');
        if(tmp > mod) break;
    }

    if(tmp <= mod) return quick_mod(a, tmp, mod);
    else{
        tmp = 0;
        for(int i = 0; i < len; i++)
            tmp = ( tmp * 10 + (b[i] - '0') ) % phi;
        return quick_mod(a, tmp + phi, mod);
    }
}

LL ans = 0;

//遞歸的求2^(2^(2^(2.....)))
/*
int dfs(int p){
    if(p == 1) return 1;

    LL phi = ouler(p);
    return quick_mod(2, dfs(phi) + phi, p);
}*/

//求a^(a^(a^(a^(a^...))))   b個a次冪
LL dfs(LL a, LL b, LL p){
    if(p == 1) return 0;
    if(b == 0) return 1;

    LL phi = ouler(p);

    LL tmp = dfs(a, b-1, phi);
    if(tmp <= phi && tmp) return quick_mod(a, tmp, p);
    else return quick_mod(a, tmp+phi, p);
}

//  A ^ B % p
int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        LL a, b, m;
        scanf("%lld%lld%lld", &a, &b, &m);
        printf("%lld\n", dfs(a, b, m) % m);
    }
    return 0;
}