题目描述：
题解：
莫比乌斯反演。
能搞出来这个式子：$\sum\limits _{T=1}^{min(n,m)} \lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor \sum\limits_{g|T} g^K \mu (\frac{T}{g})$
后面积性函数线性筛。
代码：
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int N = 10000050; const int MOD = 1000000007; template<typename T> inline void read(T&x) { T f = 1,c = 0;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){c=c*10+ch-‘0‘;ch=getchar();} x = f*c; } template<typename T>inline void Mod(T&x){if(x>=MOD)x-=MOD;} ll fastpow(ll x,int y) { ll ret = 1; while(y) { if(y&1)ret=ret*x%MOD; x=x*x%MOD;y>>=1; } return ret; } bool vis[N]; int pri[N/10],pnt,K; ll s[N]; inline void init() { s[1] = 1; for(register int i=2;i<=5000000;++i) { if(!vis[i]) { pri[++pnt]=i; s[i] = fastpow(i,K)-1; } for(register int j=1;i*pri[j]<=5000000;++j) { int now = i*pri[j]; vis[now]=1; if(i%pri[j])s[now] = s[i]*s[pri[j]]%MOD; else { s[now] = s[i]*(s[pri[j]]+1)%MOD; break; } } } for(register int i=1;i<=5000000;++i)Mod(s[i]+=s[i-1]); } int T,n,m; inline void work() { read(n),read(m); ll ans = 0;int mn = min(n,m); for(register int i=1,j;i<=mn;i=j+1) { j = min(n/(n/i),m/(m/i)); Mod(ans+=1ll*(n/i)*(m/i)%MOD*(s[j]+MOD-s[i-1])%MOD); } printf("%lld\n",ans); } int main() { freopen("tt.in","r",stdin); read(T),read(K);init(); while(T--)work(); return 0; }
原文：https://www.cnblogs.com/LiGuanlin1124/p/11149970.html
