HDU 6069 Counting Divisors

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1604    Accepted Submission(s): 592

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

 

Source

2017 Multi-University Training Contest - Team 4

 

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/*
* @Author: Lyucheng
* @Date:   2017-08-03 13:13:45
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-08-04 11:25:19
*/
/*
 题意：给你一个区间[l,r]让你求区间内每个数的k次方因子数的总和

 思路：比赛的时候想出来 i^k的因子是 (n1*k+1)*(n2*k+1)*...*(np*k+1),但是没想出来怎么优化，素数枚举
    很烦，四场比赛每次差一点，比赛的时候想的是枚举[l,r]之间的数，优化到8300ms,实在没法优化了，应
    该反过来想，枚举从[l,r]的素因子，因为i素因子的个数远小于i,大多数在sqrt i内，最多只有一个在sqrt i
    之外。
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define LL long long
#define mod 998244353
#define MAXN 1000005

using namespace std;

int t;
LL l,r,k;
LL res;
LL d[MAXN];
LL pos[MAXN];
bool prime[MAXN];
LL p[MAXN];
int tol;
void init(){
    tol=;
    for(int i=;i<MAXN;i++){
        prime[i]=true;
    }
    prime[]=prime[]=false;
    for(int i=;i<MAXN;i++){
        if(prime[i])
            p[tol++]=(LL)i;
        for(int j=;j<tol&&i*p[j]<MAXN;j++){
            prime[i*p[j]]=false;
            if((LL)i%p[j]==) break;
        }
    }
}

int main(){
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    init();
    scanf("%d",&t);
    while(t--){
        res=;
        scanf("%I64d%I64d%I64d",&l,&r,&k);
        for(LL i=l;i<=r;i++){
            d[(int)(i-l)]=;
            pos[(int)(i-l)]=i;
        }
        for(int i=;i<tol;i++){//枚举所有的素数
            if(p[i]<=) continue;
            LL cnt=(l+p[i]-)/p[i]*p[i];//找出[l,r]区间内的第一个p[i]的倍数
            if(cnt-l<||cnt-l>r-l) continue;
            for(int j=cnt-l;j<=r-l;j+=p[i]){
                LL cur=;
                while(pos[j]&&pos[j]%p[i]==){
                    cur++;
                    pos[j]/=p[i];
                }
                d[j]*=(cur*k+);
                d[j]%=mod;
            }
        }
        for(int i=;i<=r-l;i++){
            if(pos[i]==)
                res+=d[i];
            else
                res+=d[i]*(k+);
            res%=mod;
        }
        printf("%I64d\n",res);
    }
    return ;
}