求
[sum_{i=1}^n sigma(i^k)
]
我们先来设一个函数(f(i)=sigma(i^k))
根据约数个数定理
[f(p)=sigma(p^k)=k+1
]
[f(p^c)=sigma(p^{ck})=ck+1
]
这不就可以Min_25筛了吗
还是先求出来一个区间内的质数个数，一个质数的贡献显然是(k+1)，之后上板子就好了
代码
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define maxn 3000005
#define re register
#define LL unsigned long long
int T,tot,f[maxn];
LL n,K,p[maxn],w[maxn],id1[maxn],id2[maxn],m,Sqr,g[maxn];
inline LL S(LL x,int y) {
if(x<=1||p[y]>x) return 0;
int t=(x<=Sqr)?id1[x]:id2[n/x];
LL ans=g[t]*(K+1)-(K+1)*(y-1);
for(re int k=y;k<=tot&&p[k]*p[k]<=x;k++) {
LL p1=p[k];
for(re int e=1;p1<=x;e++,p1*=p[k])
ans+=(S(x/p1,k+1)+((e>1)?1:0))*(K*e+1);
}
return ans;
}
int main()
{
scanf("%d",&T);f[1]=1;
for(re int i=2;i<=100000;i++) {
if(!f[i]) p[++tot]=i;
for(re int j=1;j<=tot&&p[j]*i<=100000;j++) {
f[p[j]*i]=1;if(i%p[j]==0) break;
}
}
while(T--) {
scanf("%llu%llu",&n,&K);Sqr=std::sqrt(n)+1;m=0;
for(re LL l=1,r;l<=n;l=r+1) {
r=n/(n/l);w[++m]=n/l;
if(w[m]<=Sqr) id1[w[m]]=m;
else id2[n/w[m]]=m;
g[m]=w[m]-1;
}
for(re int j=1;j<=tot&&p[j]*p[j]<=n;j++)
for(re int i=1;i<=m&&p[j]*p[j]<=w[i];i++) {
int k=(w[i]/p[j]<=Sqr)?id1[w[i]/p[j]]:id2[n/(w[i]/p[j])];
g[i]=g[i]-g[k]+j-1;
}
printf("%llu
",S(n,1)+1);
}
return 0;
}