bzoj4407: 于神之怒加强版

题意:求\(\sum_{i=1}^n\sum_{j=1}^mgcd(i,j)^k%1e9+7\)
题解:考虑枚举gcd,原式可化简为\(\sum_{d=1}^{n}d^k\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==d]\)后面部分很明显是最基础的莫比乌斯反演,
那么有\(\sum_{d=1}^{n}d^k\sum_{x=1}^{\lfloor \frac{n}{d} \rfloor}\mu(x)*{\lfloor \frac{n}{d*x} \rfloor}*{\lfloor \frac{m}{d*x} \rfloor}\)
考虑枚举t=dx,(这里的套路很重要!!!*),那么有\(\sum_{t=1}^n{\lfloor \frac{n}{t} \rfloor}*{\lfloor \frac{m}{t} \rfloor}\sum_{d|t}d^k*\mu({\frac{t}{d}})\)
后面是一个积性函数可以O(n)预处理,前面可以分块,这里假设后面的积性函数是\(f(n)=\sum_{d|n}d^k*\mu({\frac{n}{d}})\)
\(f(n)=\prod_{i=1}^kf(p_i^{x_i})=\prod_{i=1}^k\mu(1)*p_i^{k*x_i}+\mu(p_i)*p_i^{k*(x_i-1)}=\prod_{i=1}^kp_i^{k*(x_i-1)}*(p_i^k-1)\)
这里第二步其他项没有的情况是\(\mu\)函数质因子两个以上就是0了,然后f(n)可以在线性筛的时候处理,可以先把素数的k次幂处理出来

/**************************************************************
    Problem: 4407
    User: walfy
    Language: C++
    Result: Accepted
    Time:19252 ms
    Memory:103832 kb
****************************************************************/
 
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
 
using namespace std;
 
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=5000000+10,maxn=400000+10,inf=0x3f3f3f3f;
 
int prime[N],cnt,k;
ll f[N],qk[N];
bool mark[N];
void init()
{
    f[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,qk[cnt]=qp(i,k),f[i]=(qk[cnt]-1+mod)%mod;
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                f[i*prime[j]]=f[i]*qk[j]%mod;
                break;
            }
            f[i*prime[j]]=f[i]*(qk[j]-1+mod)%mod;
        }
    }
    for(int i=1;i<N;i++)add(f[i],f[i-1]);
}
int main()
{
    int T;scanf("%d%d",&T,&k);
    init();
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        if(n>m)swap(n,m);
        ll ans=0;
        for(int i=1,j;i<=n;i=j+1)
        {
            j=min(n/(n/i),m/(m/i));
            ll te=1ll*(f[j]-f[i-1])*(n/i)%mod*(m/i)%mod;
            te=(te+mod)%mod;
            add(ans,te);
        }
        printf("%lld\n",ans);
    }
    return 0;
}
/********************
 
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