hdu6588(数论)

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=6588

题意

求 $\displaystyle \sum_{I=1}^N (\sqrt[3]i,i)\bmod{998244353}$

题解

显然枚举 $\sqrt[3]i$ ,那么有

其中 $n=\sqrt[3]N$

第二层和式可以差分化简

然后有




代码

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/**
*        ┏┓    ┏┓
*        ┏┛┗━━━━━━━┛┗━━━┓
*        ┃       ┃  
*        ┃   ━    ┃
*        ┃ >   < ┃
*        ┃       ┃
*        ┃... ⌒ ...  ┃
*        ┃       ┃
*        ┗━┓   ┏━┛
*          ┃   ┃ Code is far away from bug with the animal protecting          
*          ┃   ┃ 神兽保佑,代码无bug
*          ┃   ┃           
*          ┃   ┃       
*          ┃   ┃
*          ┃   ┃           
*          ┃   ┗━━━┓
*          ┃       ┣┓
*          ┃       ┏┛
*          ┗┓┓┏━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-12
#define succ(x) (1LL<<(x))
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid ((x+y)>>1)
#define NM 10000005
#define nm 11
#define pi 3.1415926535897931
const ll inf=998244353;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}



template <class T>
void read(T &x) {

static char ch;static bool neg;
for(ch=neg=0;ch<'0' || '9'<ch;neg|=ch=='-',ch=getchar());
for(x=0;'0'<=ch && ch<='9';(x*=10)+=ch-'0',ch=getchar());
x=neg?-x:x;
}


__int128 p;
int n,prime[NM],tot,phi[NM];
ll ans,_x,_y,pre[NM],sum[NM];
bool v[NM];

void init(){
n=1e7;phi[1]=1;
inc(i,1,n)sum[i]=(sum[i-1]+i)%inf,pre[i]=(pre[i-1]+1ll*i*i)%inf;
inc(i,2,n){
if(!v[i])prime[++tot]=i,phi[i]=i-1;
inc(j,1,tot){
if(i*prime[j]>n)break;
v[i*prime[j]]++;
if(i%prime[j]==0){
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
phi[i*prime[j]]=phi[i]*phi[prime[j]];
}
}
}

int main(){
init();
int _=read();while(_--){
read(p);
for(__int128 x=0,y=1e7;x<=y;){
__int128 t=x+y>>1;
if(t*t*t<=p)n=t,x=t+1;else y=t-1;
}
n--;ans=1ll*n*(n+1)/2%inf;
for(int i=1;i<=n;i++){
ans+=3ll*phi[i]*i%inf*pre[n/i]%inf;
ans+=3ll*phi[i]*sum[n/i]%inf;
ans%=inf;
}
n++;
__int128 t=n;t=t*t*t-1;
for(int i=1;i*i<=n;i++)if(n%i==0){
ans+=phi[i]*(p/i%inf-t/i%inf+inf)%inf;
if(i*i<n)ans+=phi[n/i]*(p/(n/i)%inf-t/(n/i)%inf+inf)%inf;
ans%=inf;
}
printf("%lld\n",ans);
}
return 0;
}