hdu 4910 Problem about GCD(數論)，hdu4910

hdu 4910 Problem about GCD(數論)，hdu4910

題目串連：hdu 4910 Problem about GCD

題目大意：給定M，判斷所有小於M並且和M互質的數的積模數M的值。

解題思路：有個數論的結論，若為偶數，M=M/2. 可以寫成M=pk,即只有一種質因子時，答案為M-1，否則為1.特殊情況為4的倍數，不包括4.
首先用1e6以內的素數去試除，如果都不可以為p，那麼對大於1e6的情況判斷一下是否為素數，是素數也可以（k=1），否則開方計算，因為M最大為1e18，不可能包含3個大於1e6的質因子。

#include <cstdio>#include <cstring>#include <cmath>#include <ctime> // possible need;#include <cstdlib> // possible need;#include <iostream>#include <algorithm>using namespace std;typedef long long type;type mul_mod (type a, type b, type mod) {    type ret = 0;    while (b) {        if (b&1)            ret = (ret + a) % mod;        a = (a + a) % mod;        b >>= 1;    }    return ret;}type pow_mod (type a, type n, type mod) {    type ans = 1;    while (n) {        if (n&1)            ans = mul_mod(ans, a, mod);        a = mul_mod(a, a, mod);        n >>= 1;    }    return ans;}bool miller_rabin(type n) {    if (n < 2)        return false;    srand(time(0));    for (int i = 0; i < 20; i++)        if (pow_mod(rand() % (n-1) + 1, n-1, n) != 1)            return false;    return true;}typedef long long ll;const int maxn = 1e6;bool vis[maxn+5];int np, pri[maxn+5];void prime_table (int n) {    np = 0;    memset(vis, 0, sizeof(vis));    for (int i = 3; i <= n; i += 2) {        if (vis[i])            continue;        pri[np++] = i;        for (int j = i * 2; j <= n; j += i)            vis[j] = 1;    }}bool judge (ll M) {    if (M % 2 == 0)        M /= 2;    for (int i = 0; i < np; i++) {        ll tmp = M;        while (tmp % pri[i] == 0)            tmp /= pri[i];        if (tmp == 1)            return true;    }    if (M <= 1000000LL)        return false;    if (miller_rabin(M))        return true;    ll x = sqrt(M+0.0);    while (x * x < M)        x++;    if (x * x != M)        return false;    if (miller_rabin(x))        return true;    return false;}int main () {    ll M;    prime_table(maxn);    //srand(time(0));    while (scanf("%I64d", &M) == 1 && M != -1) {        if (M == 1 || M == 2 || M == 4 || judge(M))            printf("%I64d\n", M-1);        else            printf("1\n");    }    return 0;}