HDU 6706 2019CCPC网络赛 1005 huntian oy（杜教筛）-CFANZ编程社区

HDU 6706 2019CCPC网络赛 1005 huntian oy（杜教筛）_2019CCPC网络赛

HDU 6706 2019CCPC网络赛 1005 huntian oy（杜教筛）_2019CCPC网络赛_02

大致题意：告诉你n、a和b，让你求

$\large f(n,a,b)=\sum_{i=1}^n \sum_{j=1}^i gcd(i^a-j^a,i^b-j^b)[gcd(i,j)=1]\%(10^9+7)$

中间那个gcd(i^a-j^a,i^b-j^b)这个看起来很吓人，其实你打表会发现其实它等于i-j，那么就变成了求

$\large f(n,a,b)=\sum_{i=1}^n \sum_{j=1}^i (i-j)[gcd(i,j)=1]\%(10^9+7)$

这个显然可以拆成两个部分

$\large f(n,a,b)=\sum_{i=1}^n \sum_{j=1}^i i[gcd(i,j)=1] -\sum_{i=1}^n \sum_{j=1}^ij[gcd(i,j)=1]$

第一部分就是phi(i)的和，第二部分就是所有与i互质的小于i的数字的和，即

$\large f(n,a,b)=\sum_{i=1}^n i*\phi(i) -\sum_{i=1}^n \frac{i*\phi(i)+[i=1]}{2}=\sum_{i=1}^n \frac{i*\phi(i)-1}{2}$

利用杜教筛尝试求i*phi(i)的前缀和，我们令f(i)=i*phi(i)，有

$\large \begin{align*} f(i)&=i*\phi(i) \\ (f*id)(i)&=\sum_{i|j}i/j*\phi(i/j)*j\\ (f*id)(i)&=i*\sum_{i|j}\phi(i/j)\\ (f*id)(i)&=i^2\end{align*}$

然后考虑开始求和

$\large \begin{align*} \sum_{i=1}^{n} (f*id)(i)&=\sum_{i=1}^{n}i^2\\ \sum_{i=1}^{n}\sum_{i|j}f(i/j)*id(j)&=\frac{n(n+1)(2n+1)}{6}\\ \sum_{j=1}^{n}j\sum_{i=1}^{\lfloor\frac{n}{j}\rfloor}f(i)&=\frac{n(n+1)(2n+1)}{6}\\ \sum_{j=1}^{n}j*S(\lfloor n/j \rfloor)&=\frac{n(n+1)(2n+1)}{6}\\ S(n)&=\frac{n(n+1)(2n+1)}{6}-\sum_{j=2}^{n}j*S(\lfloor n/j \rfloor) \end{align*}$

套上杜教筛即可，具体见代码：

#include<bits/stdc++.h>
#define N 3333310
#define INF 0x3f3f3f3f
#define eps 1e-6
#define pi 3.141592653589793
#define mod 1000000007
#define P 1000000007
#define LL long long
#define pb push_back
#define fi first
#define se second
#define cl clear
#define si size
#define lb lower_bound
#define ub upper_bound
#define bug(x) cerr<<#x<<"      :   "<<x<<endl
#define mem(x) memset(x,0,sizeof x)
#define sc(x) scanf("%d",&x)
#define scc(x,y) scanf("%d%d",&x,&y)
#define sccc(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace __gnu_pbds;
unordered_map<LL,LL> mp;

const LL inv2=500000004;
const LL inv6=166666668;

int p[N],cnt,phi[N],psum[N],isp[N];
int vis[N];

void init()
{
    int sz=0;
    phi[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!isp[i])p[++sz]=i,phi[i]=i-1;
         for(int j=1;j<=sz&&(LL)i*p[j]<N;j++)
         {
            isp[i*p[j]]=1;
            if(i%p[j]==0)
            {
                phi[p[j]*i]=phi[i]*p[j];
                break;
            } else phi[p[j]*i]=(p[j]-1)*phi[i];
        }
    }
    for(int i=1;i<N;i++)
        psum[i]=(psum[i-1]+(LL)i*phi[i]%mod)%mod;
}

LL S(LL x) {
    if(mp[x]) return mp[x];
    if(x<N) return psum[x];
    LL ans=0,res=((x*(x+1)%mod)*((2*x+1)%mod))%mod;
    res=res*inv6%mod;
    for(LL l=2,r;l<=x;l=r+1) {
        LL tmp=x/l; r=x/tmp;
        ans=(ans+((r+l)%mod*(r-l+1+mod)%mod*inv2%mod)*S(tmp)%mod)%mod;
    }
    return mp[x]=(res-ans+mod)%mod;
}

int main() {
    init();
    int T;
    for(sc(T);T;T--){
        LL n,a,b,ans;
        sccc(n,a,b);
        ans=(S(n)-1+mod)%mod;
        ans=ans*inv2%mod;
        printf("%lld\n",ans%mod);
    }
    return 0;
}