P4449 于神之怒加強版

推式子

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

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

$$=\sum _{d=1}^n{d}^h\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}\sum _{k\mid gc(i,j)}\mu (k)$$

$$=\sum _{d=1}^n{d}^h\sum _{k=1}^{\frac{n}{d}}\mu (k)\sum _{i=1}^{\frac{n}{dk}}\sum _{j=1}^{\frac{m}{dk}}$$

$$=\sum _{d=1}^n{d}^h\sum _{k=1}^{\frac{n}{d}}\mu (k)?\frac{n}{kd}??\frac{m}{kd}?$$

$t=kd$
$$=\sum _{t=1}^n?\frac{n}{t}??\frac{m}{t}?\sum _{d\mid t}{d}^h\mu (\frac{t}{d})$$
所以我們只要再預處理出$f(t)={\sum }_{d\mid t}{d}^h\mu (\frac{t}{d})$即可，由于后面是一個積性函數我們可以直接通過素數篩來得到。

代碼

/*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' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using 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;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }const int N = 5e6 + 10, mod = 1e9 + 7;bool st[N];ll mu[N], n, m, k, sum[N], f[N];int prime[N], tot;ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }void mobius() {sum[1] = 1;for (int i = 2; i < N; i++) {if(!st[i]){prime[++tot] = i;f[tot] = quick_pow(i, k, mod);sum[i] = (f[tot] + mod - 1) % mod;}for(int j = 1; j <= tot && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {sum[i * prime[j]] = sum[i] * f[j] % mod;break;}sum[i * prime[j]] = sum[i] * sum[prime[j]] % mod;}}for(int i = 2; i < N; i++) {sum[i] = (sum[i - 1] + sum[i]) % mod;} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = read();k = read();mobius();while(T--) {n = read(), m = read();if(n > m) swap(n, m);ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans = (ans + (n / l) * (m / l) % mod * (sum[r] - sum[l - 1] + mod) % mod + mod) % mod;}printf("%lld\n", ans);}return 0; }

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