题意:提前给出\(k\),求\(\sum\limits_{i=1}^n \sum\limits_{j=1}^m gcd(i,j)^k\)
套路推♂倒
\[ \sum_{D=1}^n \sum_{d|D} d^k\mu(\frac{D}{d}) \frac{n}{D} \frac{m}{D} \]
是一个\(g = idk * \mu\)啊,单位幂函数和莫比乌斯函数的卷积!
\(g(1) = 1\)
\(g(p) = -1 + p^k\)
因为带着\(\mu\),只有sf才有贡献
所以\(p \mid i\)只能把\(p\)放到\(d^k\)里了,就是\(g(i)\cdot p\)
或者考虑\(g(p^k)\)只有1和p有贡献也可以,直接得到计算公式再递推
#include
#include
#include
#include
#include
using namespace std;
const int N=5e6+5, INF=1e9, P=1e9+7;
#define pii pair
#define MP make_pair
#define fir first
#define sec second
typedef long long ll;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
int n, m, k;
int notp[N], p[N], mu[N];
ll pk[N], g[N];
inline ll Pow(ll a, int b) {
ll ans = 1;
for(; b; b>>=1, a=a*a%P)
if(b&1) ans = ans*a%P;
return ans;
}
inline ll powk(int a) {return pk[a] ? pk[a] : pk[a]=Pow(a, k);}
void sieve(int n) {
mu[1] = 1; g[1] = 1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, mu[i] = -1, g[i] = -1 + powk(i);
for(int j=1; j<=p[0] && i*p[j]<=n; j++) {
int t = i*p[j];
notp[t] = 1;
if(i%p[j] == 0) {
g[t] = g[i]*powk(p[j])%P;
mu[t] = 0;
break;
}
g[t] = g[i]*g[p[j]]%P;
mu[t] = -mu[i];
}
}
for(int i=1; i<=n; i++) g[i] = (g[i] + g[i-1])%P;
}
ll cal(int n, int m) {
ll ans=0; int r;
for(int i=1; i<=n; i=r+1) {
r = min(n/(n/i), m/(m/i));
ans = (ans+ (g[r] - g[i-1]) * (n/i)%P * (m/i)%P )%P;
}
return (ans + P) %P;
}
int main() {
//freopen("in","r",stdin);
int T=read(); k=read();
sieve(N-1);
while(T--) {
n=read(); m=read();
if(n>m) swap(n, m);
printf("%lld\n", cal(n, m));
}
}