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UVA10006 - Carmichael Numbers(筛选构造素数表+快速幂)

UVA10006 - Carmichael Numbers(筛选构造素数表+快速幂)

题目链接

题目大意:如果有一个合数,然后它满足任意大于1小于n的整数a, 满足a^n%n = a;这样的合数叫做Carmichael Numbers。题目给你n,然你判断是不是Carmichael Numbers。

解题思路:首先用筛选法构造素数表,判断n是否是合数,然后在用快速幂求a^2-a^(n - 1)是否满足上述的式子。快速幂的时候最好用long long ,防止相乘溢出。

代码:

#include <cstdio>
#include <cstring>
#include <cmath>

const int maxn = 65000 + 5;
typedef long long ll;

int notprime[maxn];

void init () {

    for (int i = 2; i < maxn; i++) 
        for (int j = 2 * i; j < maxn; j += i) 
            notprime[j] = 1;
}

ll powmod(ll x, ll n, ll mod) {

    if (n == 1)
        return x;

    ll ans = powmod(x, n / 2, mod);
    ans = (ans * ans) % mod;
    if (n % 2 == 1)
        ans *= x;
    return ans % mod;    
}

bool is_carmichael(int n) {

    for (int i = 2; i < n; i++) {
        if (powmod(i, n, n) != i)
        return false;
    }
    return true;
}

int main () {


    init();
    int n;
    while (scanf ("%d", &n) && n) {

        if (notprime[n] == 0)
            printf ("%d is normal.\n", n);
        else {
            bool flag = is_carmichael(n);
            if (flag)
                printf ("The number %d is a Carmichael number.\n", n);
            else
                printf ("%d is normal.\n", n);
        }
    }
    return 0;
}

UVA10006 - Carmichael Numbers(筛选构造素数表+快速幂)