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)：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<stdio.h>
 6 #define it (p-l)
 7 using namespace std;
 8 typedef long long LL;
 9 const LL MOD=998244353;
10 const long long MAXN=1000005;
11 long long prime[MAXN],tot=0;
12 bool isPrime[MAXN];
13 LL k,num[MAXN],res[MAXN];
14 void getprime(){
15     memset(isPrime, true, sizeof(isPrime));
16     for(int i=2;i<MAXN;i++){
17         if(isPrime[i]){
18             prime[++tot]=i;
19         }
20         for(int j=1;j<=tot;j++){
21             if(i*prime[j]>MAXN) break;
22                 isPrime[i*prime[j]]=false;
23             if(i%prime[j]==0) break;
24         }
25     }
26     return ;
27 }
28 LL cal(LL l, LL r)
29 {
30     LL ans=0,tmp,cnt;
31     for(int i=1;i<=tot;i++)
32     {
33         LL p=(l+prime[i]-1)/prime[i]*prime[i];
34         while(p<=r){
35             cnt=0;
36             while(num[it]%prime[i]==0){
37                 num[it]/=prime[i];
38                 cnt++;
39             }
40             res[it]=res[it]*(k*cnt+1)%MOD;
41             p+=prime[i];
42         }
43     }
44     for(LL p=l;p<=r;p++){
45         if(num[it]==1)
46             ans+=res[it];
47         else
48             ans+=res[it]*(k+1);
49         ans%=MOD;
50     }
51     return ans;
52 }
53 int main()
54 {
55     int T;
56     LL l,r;
57     getprime();
58     scanf("%d", &T);
59     while(T--)
60     {
61         scanf("%lld %lld %lld", &l, &r, &k);
62         for(LL p=l;p<=r;p++){
63             res[it]=1;
64             num[it]=p;
65         }
66         LL res=cal(l, r);
67         printf("%lld\n", res);
68     }
69 }