UVA10515 Powers Et Al（欧拉降幂）

Description：

Finding the exponent of any number can be very troublesome as it grows exponentially. But in this problem you will have to do a very simple task. Given two non-negative numbers $m$ and $n$ , you will have to find the last digit of m^n in decimal number system.

Input

The input file contains less than 100000 lines. Each line contains two integers $m$ and $n$ (Less than $10^{101}$ . Input is terminated by a line containing two zeroes. This line should not be processed.

Output

For each set of input you must produce one line of output which contains a single digit. This digit is the last digit of $m^n$ .

Sample Input

2 2

2 5

0 0

Sample Output

题意：

求 $m^n$ 的最后一位，数很大套欧拉降幂模板，把模数改为 $10$ .

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", 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;
}

const int maxn = 1e6 + 10;
char b[maxn], s[maxn];
ll c;
ll mult(ll a, ll b) ///a^b%c
{
	ll z = 1;
	while (b)
	{
		if (b & 1)
			z = (z * a) % c;
		a = (a * a) % c;
		b /= 2;
	}
	return z;
}
ll phi(ll n) ///欧拉函数
{
	ll rea = n;
	for (int i = 2; i * i <= n; i++)
		if (n % i == 0)
		{
			rea = rea - rea / i;
			do
				n /= i;
			while (n % i == 0);
		}
	if (n > 1)
		rea = rea - rea / n;
	return rea;
}
int main()
{
	ll bb, n, len1, len2;
	while (~scanf("%s%s", b, s))
	{
		c = 10;
		bb = 0;
		n = 0;
		len1 = strlen(b);
		len2 = strlen(s);
		if (len1 == 1 && len2 == 1 && b[0] == '0' && s[0] == '0')
		{
			break;
		}
		int cnt = phi(10);
		rep(i, 0, len1 - 1)
		{
			bb = (bb * 10 + b[i] - '0') % c;
		}
		bool ok = 0;
		rep(i, 0, len2 - 1)
		{
			n = n * 10 + s[i] - '0';
			if (n > cnt)
			{
				ok = 1;
				n = n % cnt;
			}
		}
		if (ok)
			n += cnt;
		ll ans = 0;
		if (bb == 0)
			bb = c;
		ans = mult(bb, n);
		pld(ans);
	}
	return 0;
}

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