P4449 于神之怒加强版
推式子
∑i=1n∑j=1ngcd(i,j)h\sum_{i = 1} ^{n} \sum_{j = 1} ^{n} gcd(i, j) ^ h i=1∑nj=1∑ngcd(i,j)h
=∑d=1ndh∑i=1nd∑j=1mdgcd(i,j)==1=\sum_{d = 1} ^{n} d ^ h \sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}}gcd(i, j) == 1 =d=1∑ndhi=1∑dnj=1∑dmgcd(i,j)==1
=∑d=1ndh∑i=1nd∑j=1md∑k∣gc(i,j)μ(k)=\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) =d=1∑ndhi=1∑dnj=1∑dmk∣gc(i,j)∑μ(k)
=∑d=1ndh∑k=1ndμ(k)∑i=1ndk∑j=1mdk=\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}} =d=1∑ndhk=1∑dnμ(k)i=1∑dknj=1∑dkm
=∑d=1ndh∑k=1ndμ(k)⌊nkd⌋⌊mkd⌋=\sum_{d = 1} ^{n} d ^ h \sum_{k = 1} ^{\frac{n}{d}} \mu(k) \lfloor\frac{n}{kd}\rfloor \lfloor\frac{m}{kd}\rfloor =d=1∑ndhk=1∑dnμ(k)⌊kdn⌋⌊kdm⌋
t=kdt = kdt=kd
=∑t=1n⌊nt⌋⌊mt⌋∑d∣tdhμ(td)=\sum_{t = 1} ^{n} \lfloor \frac{n}{t} \rfloor \lfloor \frac{m}{t} \rfloor \sum_{d \mid t} d ^ h \mu(\frac{t}{d}) =t=1∑n⌊tn⌋⌊tm⌋d∣t∑dhμ(dt)
所以我们只要再预处理出f(t)=∑d∣tdhμ(td)f(t) = \sum_{d \mid t} d ^ h \mu(\frac{t}{d})f(t)=∑d∣tdhμ(dt)即可,由于后面是一个积性函数我们可以直接通过素数筛来得到。
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'
#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;
}