HDU 2837 Calculation（欧拉定理+递归）

Description：

Assume that $f(0) = 1$ and $0^0=1$ . $f(n) = (n\%10)^{f(n/10)}$ for all n bigger than zero. Please calculate $f(n)\%m. (2 ≤ n , m ≤ 10^9, x^y$ means the $y$ th power of $x$ ).

Input

The first line contains a single positive integer $T$ . which is the number of test cases. T lines follows.Each case consists of one line containing two positive integers $n$ and $m$ .

Output

One integer indicating the value of $f(n)\%m$ .

Sample Input

2
24 20
25 20

Sample Output

16
5

题意：

给出 $f(0) = 1$ $f(n) = (n\%10)^{f(n/10)}$ 求 $f(n)\%m$ ，写几项就会发现规律。

比如 $f(45)=5^4$ $f(125)=5^{2^1}$ $f(4567)=7^{6^{5^4}}$ ，自己一开始直接使用的快速幂求

解，一直错，然后是因为这个指数太大，存在循环节，可以使用欧拉函数降幂。

$a^x=a^{x\%phi(c)+phi(c)} (mod c)$

然后再递归求解，注意使用快速幂过程中需要对底数和模数的关系进行判断。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
		a = a % mod + mod;
	ll ans = 1;
	while (b)
	{
		if (b & 1)
		{
			ans = ans * a;
			if (ans >= mod)
				ans = ans % mod + mod;
		}
		a *= a;
		if (a >= mod)
			a = a % mod + mod;
		b >>= 1;
	}
	return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

ll n, m;
int t;
ll ans;


ll phi(ll n)
{
    ll i, rea = n;
    for (i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            rea = rea - rea / i;
            while (n % i == 0)
                n /= i;
        }
    }
    if (n > 1)
        rea = rea - rea / n;
    return rea;
}

ll dfs(ll n, ll m)
{
    if (n == 0)
        return 1;
    if (n < 10)
        return n;
    ll res = phi(m);
    ll temp = dfs(n / 10, res);
    return qpow(n % 10, temp, m);
}

int main()
{
    sd(t);
    while (t--)
    {
        sldd(n, m);
        ans = dfs(n, m) % m;
        pld(ans);
    }
    return 0;
}

来源：CSDN

作者：邵光亮

链接：https://blog.csdn.net/qq_43627087/article/details/103976628

标签

欧拉定理

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