HDU 6069 Counting Divisors —— 2017 Multi-University Training 4

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2599    Accepted Submission(s): 959

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 :

 

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

 

题目大意：d（i）是 i 的因数个数，让我们求 l<=i<=r 时，d（i^k）之和.

思路：对一个数n=p₁^t₁*p₂^t₂*...*p_n^t_n, p_i是n的质因数。则n的因数个数是(t₁+1)*(t₂+1)*...*(t_n-1+1)*(t_n+1), 易得i^k的因数个数是(k*t₁+1)*(k*t₂+1)*...*(k*t_n+1)，那么接下来就是要对 i 进行质因数分解了。 在打表打出质因数后，分解时对于一个质数P, 在[l , r]区间内所有能整除P的数进行质因数分解，这样能保证不会有多余时间花在搜索质因数上，这种做法类似筛法。具体见代码。

 

AC代码(标程)：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<stdio.h>
#define it (p-l) 
using namespace std;
typedef long long LL;
const LL MOD=998244353;
const long long MAXN=1000005;
long long prime[MAXN],tot=0;
bool isPrime[MAXN];    
LL k,num[MAXN],res[MAXN];
void getprime(){
    memset(isPrime, true, sizeof(isPrime));
    for(int i=2;i<MAXN;i++){
        if(isPrime[i]){
            prime[++tot]=i;
        }
        for(int j=1;j<=tot;j++){
            if(i*prime[j]>MAXN) break;
                isPrime[i*prime[j]]=false;
            if(i%prime[j]==0) break;
        }
    }
    return ;
}
LL cal(LL l, LL r)
{
    LL ans=0,tmp,cnt;
    for(int i=1;i<=tot;i++)
    {
        LL p=(l+prime[i]-1)/prime[i]*prime[i];
        while(p<=r){
            cnt=0;
            while(num[it]%prime[i]==0){
                num[it]/=prime[i];
                cnt++;
            }
            res[it]=res[it]*(k*cnt+1)%MOD;
            p+=prime[i];
        }
    }
    for(LL p=l;p<=r;p++){
        if(num[it]==1)
            ans+=res[it];
        else
            ans+=res[it]*(k+1);
        ans%=MOD;
    }
    return ans;
}
int main()
{
    int T;
    LL l,r;
    getprime();
    scanf("%d", &T);
    while(T--)
    {
        scanf("%lld %lld %lld", &l, &r, &k);
        for(LL p=l;p<=r;p++){
            res[it]=1;
            num[it]=p;
        }
        LL res=cal(l, r);
        printf("%lld
", res);
    }
}