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

3
1 5 1
1 10 2
1 100 3

Sample Output

10
48
2302

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 long
using namespace std;
#define ll long long
const 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;
}