hdu 2685 I won't tell you this is about number theory 数论

题目链接

根据公式

\[gcd(a^m-1, a^n-1) = a^{gcd(m, n)}-1
\]

就可以很容易的做出来了。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int gcd(int a, int b) {
    return b?gcd(b, a%b):a;
}
int pow(int a, int b, int k) {
    int tmp = 1;
    while(b) {
        if(b&1) {
            tmp = tmp*a%k;
        }
        a = a*a%k;
        b>>=1;
    }
    return tmp;
}
int main()
{
    int t;
    cin>>t;
    while(t--) {
        int a, k, m, n;
        scanf("%d%d%d%d", &a, &m, &n, &k);
        int tmp = gcd(m, n);
        tmp = pow(a, tmp, k);
        tmp = (tmp-1+k)%k;
        printf("%d\n", tmp);
    }
    return 0;
}