求大数的n次方对m取模（欧拉降幂）

求2^n次方对mod取模，n为一个大数

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
char s[MAXN];

long long phi(long long x)
{
	long long ret = x;
	for (int i = 2; i * i <= x; ++i)
		if (x % i == 0)
		{
			ret -= ret / i;
			while (x % i == 0)
				x /= i;
		}
	if (x > 1)
		ret -= ret / x;
	return ret;
}
ll mpow(ll a, ll n, ll m) //a ^ n % m
{
	ll t = 1;
	while (n)
	{
		if (n & 1)
			t = (t * a) % m;
		a = (a * a) % m, n >>= 1;
	}
	return t;
}
int main()
{
#ifdef LOCAL
	//freopen("C:/input.txt", "r", stdin);
#endif
	int T;
	cin >> T;
	while (T--)
	{
		ll n = 0, p = phi(MOD);
		scanf("%s", s);		//求2^s次方  s为一个大数
		for (int i = 0; s[i]; i++)
			n = (n * 10 + s[i] - '0') % p;
		n += p - 1;
		ll ans = mpow(2, n, MOD);
		cout << ans << endl;
	}

	return 0;
}