POJ - 1286 Necklace of Beads——Polya

裸的Polya

注意n可能是0，不处理的话最后/0会RE

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) { return (b == 0 ? a : gcd(b, a % b)); }
ll ipow(ll a, int n) {
    ll ans = 1;
    while (n) {
        if (n & 1) ans *= a;
        a *= a;
        n >>= 1;
    }
    return ans;
}
int main() {
    int n;
    while (~scanf("%d", &n) && ~n) {
        if (n == 0) {
            printf("0\n");
            continue;
        }
        ll ans = 0;
        for (int i = 0; i < n; i++) ans += ipow(3, gcd(i, n));
        if (n & 1) ans += n * ipow(3, (n+1)/2);
        else {
            ans += n/2 * ipow(3, (n+2)/2);
            ans += n/2 * ipow(3, n/2);
        }
        ans /= (2 * n);
        printf("%lld\n", ans);
    }
    return 0;
}