super a ^ b mod c问题_hyz b-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43537070/article/details/103221881

$super\ a^b\ mod\ c$

题目大意：

给定一个整型a, 数组b, 模数c, 求 $a^b\ mod\ c$

考察对欧拉定理的降幂应用，直接将每一位都进行取模即可

代码如下:

#include <cstdio>
#include <cstring>
#include <cctype>
#define pk putchar(' ')
#define ph puts("")
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x)
{
    x = 0;
    int f = 1;
    char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    x *= f;
}
template <class T>
void pt(T x)
{
    if (x < 0)
        putchar('-'), x = (~x) + 1;
    if (x > 9)
        pt(x / 10);
    putchar(x % 10 ^ 48);
}
int phi(int x)
{
    int res = x;
    for (int i = 2;i * i <= x; i++)
        if (x % i == 0)
        {
            res = res / i * (i - 1);
            while (x % i == 0) x /= i;
        }
    if (x > 1)
        res = res / x * (x - 1);
    return res;
}
const int N = 1e6 + 5;
int a, mod;
char b[N];
int qkpow(int a, ll b)
{
    int res = 1;
    while (b)
    {
        if (b & 1)
            res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        b >>= 1;
    }
    return res;
}
int main()
{
    while (scanf ("%d %s %d", &a, b, &mod) != EOF)
    {
        int i, len = strlen(b), pc = phi(mod);
        ll now = 0;
        for (i = 0;i < len; i++)
        {
            now = now * 10 + b[i] - 48;
            if (now > pc)
                break;
        }
        if (i == len)
            pt(qkpow(a, now)), ph;
        else
        {
            now = 0;
            for (int i = 0;i < len; i++)
                now = (now * 10 + b[i] - 48) % mod;
            pt(qkpow(a, now + pc)), ph;
        }
    }
    return 0;
}