Goffi and GCD

思路

题目要求${\sum }_{i=1}^n{\sum }_{j=1}^{n}gcd(n-i,n)gcd(n-j,n)==n^k$

显然有$gcd(n-i,n)<=n$对于$k>=3$直接可以特判$0$，对于$k==2$的时候也可以特判一定是$gcd(0,n)gcd(0,n)=n^2$。

所以我们只要考虑$k==1$的情况：

$$\sum _{i=1}^n\sum _{j=1}^{n}gcd(n-i,n)gcd(n-j,n)==n^k$$

$$=\sum _{d\mid n}\sum _{i=1}^n\sum _{i=1}^n(gcd(n-i,d)==d)(gcd(n-j,\frac{n}{d})==\frac{n}{d})$$

$$=\sum _{d\mid n}\sum _{i=1}^{\frac{n}{d}}\sum _{i=1}^d(gcd(i,d)==1)(gcd(j,\frac{n}{d})==1)$$

$$=\sum _{d\mid n}\varphi (d)\varphi (\frac{n}{d})$$

最后再特判一下$n==1$的时候是一定有一个答案的。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;ll eular(ll x) {ll ans = x;for(ll i = 2; i * i <= x; i++) {if(x % i == 0) {while(x % i == 0) {x /= i;}ans = ans / i * (i - 1);}}if(x != 1) ans = ans / x * (x - 1);return ans;
}const int mod = 1e9 + 7;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll n, k;while (scanf("%lld %lld", &n, &k) != EOF) {ll ans = 0;if (k == 2 || n == 1) ans = 1;else if (k == 1) {for (ll i = 1; i * i <=  n; i++) {if (n % i == 0) {if (i * i != n) ans = (ans + eular(n/i) * eular(i) * 2) % mod;else ans = ans = (ans + eular(i) * eular(i)) % mod;}}}printf("%lld\n", ans);}return 0;
}