给定n,m,k,计算\(\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^k\)对1000000007取模的结果
多组数据，T<=2000,1<=N,M,K<=5000000
推式子
\(\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^k\)
\(=\sum_{p=1}^np^k\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=p]\)
\(=\sum_{p=1}^np^{k}\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}[\gcd(i,j)=1]\)
\(=\sum_{p=1}^np^{k}\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}\sum_{d|i,d|j}\mu(d)\)
\(=\sum_{p=1}^np^{k}\sum_{d=1}^n\mu(d)\lfloor\frac n{dp}\rfloor\lfloor\frac m{dp}\rfloor\)
\(=\sum_{q=1}^n\sum_{p|q}p^{k}\mu(\frac qp)\lfloor\frac n{q}\rfloor\lfloor\frac m{q}\rfloor\)
注意这里求得是个数，不需要提出\(p^2\)和\(d^2\)，我式子推错了两次。。。
还是枚举倍数对于所有q处理\(\sum_{p|q}p^{k}\mu(\frac qp)\)，然后打数论分块
注意这里如果定义p为1e9+7就不要再用p了。。。
#include <cstdio>
#include <functional>
using namespace std;
int n, prime[5000010], mu[5000010], tot, fuck = 5000000, p = 1000000007;
int s[5000010];
bool vis[5000010];
int qpow(int x, int y)
{
int res = 1;
while (y > 0)
{
if (y & 1) res = res * (long long)x % p;
x = x * (long long)x % p;
y >>= 1;
}
return res;
}
int main()
{
int t, k; scanf("%d%d", &t, &k);
mu[1] = 1;
for (int i = 2; i <= fuck; i++)
{
if (vis[i] == 0) prime[++tot] = i, mu[i] = -1;
for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
{
vis[i * prime[j]] = true;
if (i % prime[j] == 0) break;
mu[i * prime[j]] = -mu[i];
}
}
for (int pp = 1; pp <= fuck; pp++)
{
int sb = qpow(pp, k);
for (int q = pp, d = 1; q <= fuck; q += pp, d++)
s[q] = (s[q] + sb * mu[d]) % p;
}
for (int i = 1; i <= fuck; i++)
{
// printf("s[%d] = %d\n", i, s[i]);
s[i] = (s[i] + s[i - 1]) % p;
}
while (t --> 0)
{
int n, m, ans = 0;
scanf("%d%d", &n, &m); if (n > m) swap(n, m);
for (int i = 1, j; i <= n; i = j + 1)
j = min(n / (n / i), m / (m / i)), ans = (ans + (s[j] - s[i - 1]) * (long long)(n / i) % p * (m / i) % p) % p;
if (ans < 0) ans += p;
printf("%d\n", ans);
}
return 0;
}56行，交上去一遍A