HDU 6069

Counting Divisors

Problem Description

In mathematics, the function d(n) denotes the number of divisors of positive integer n.

For example, d(12)=6 because 1,2,3,4,6,12 are all 12's divisors.

In this problem, given l,r and k, your task is to calculate the following thing :

(∑i=lrd(ik))mod998244353

 

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 3 integers l,r,k(1≤l≤r≤1012,r−l≤106,1≤k≤107).

 

 

Output

For each test case, print a single line containing an integer, denoting the answer.

 

 

Sample Input

31 5 11 10 21 100 3

 

 

Sample Output

10482302

 

这题实质上就是分解质因数，不过不能对每个数都分解一次，这样肯定超时。

要用线性的方法求质因数。

设i可以分解为a1,a2,a3,a4……am,则总数加上（a1*k+1)*(a2*k+1)*……(am*k+1)

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#define ll long longusing namespace std;#define ll long longconst int mod=998244353;const int maxn=1000005;int prime[maxn];bool vis[maxn];int top;ll a[maxn];ll b[maxn];void pri(){    top=0;    memset(vis,0,sizeof vis);    vis[1]=1;    for(int i=2; i<maxn; i++)    {        if(!vis[i])            prime[top++]=i;        for(int j=0; j<top&&i*prime[j]<maxn; j++)        {            vis[i*prime[j]]=1;            if(i%prime[j]==0)                break;        }    }}void fun(ll l,ll r,ll k){    for(ll i=l; i<=r; i++)        b[i-l]=i;    for(ll i=l; i<=r; i++)        a[i-l]=1;    for(ll i=0; i<top&&prime[i]<=sqrt(r); i++)    {        ll x=l/prime[i];        if(x*prime[i]<l)            x++;        for(ll j=x; j*prime[i]<=r; j++)        {            ll s=0;            while(b[prime[i]*j-l]%prime[i]==0)            {                s++;                b[prime[i]*j-l]/=prime[i];            }            a[prime[i]*j-l]=a[prime[i]*j-l]*(s*k+1)%mod;        }    }    for(ll i=l; i<=r; i++)        if(b[i-l]>1)            a[i-l]=a[i-l]*(k+1)%mod;}int main(){    pri();    int T;    scanf("%d",&T);    while(T--)    {        ll l,r;        ll k;        scanf("%lld%lld%lld",&l,&r,&k);        ll sum=0;        fun(l,r,k);        for(ll i=l; i<=r; i++)            sum=(sum+a[i-l])%mod;        printf("%lld\n",sum);    }    return 0;}

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