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
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 #include
          
            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) 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 else48 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 }