多校的1003

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

31 5 11 10 21 100 3

 

Sample Output

这道没写出来，不过用质数求约数的个数，用了快速幂的方法

#include<iostream>#include<map>#include <cstdio>#define mod 998244353#define ll long longusing namespace std;ll prime[100005]={2},P=1,N,Num;map<ll,ll> divisor;map<ll,ll>::iterator it;ll isPrime(ll N){      if(N % 3==0) return N==3;      for(ll i=1;prime[i]*prime[i]<=N;++i)            if( N % prime[i]==0) return 0;      return 1;}ll divNum(ll Num){      for(ll i=0;prime[i]<=Num && i<P;++i)             while( Num % prime[i] == 0 ){                     divisor[prime[i]]++;                     Num /= prime[i];             }      if(Num!=1) divisor[Num]=1;      int result=1;      for(it=divisor.begin();it!=divisor.end();++it)            result *= (it->second + 1);      divisor.clear();      return result;}ll quickpow(ll a, ll b){    ll ans=1, base=a;    while(b!=0){        if(b&1!=0)            ans*=base%mod;        base*=base%mod;        b>>=1;    }    return ans%mod;}int main(){     ll i;     for(i=3;i<=1000000;i+=2)           if(isPrime(i)) {            prime[P++]=i;           }     int tcase;     scanf("%d",&tcase);     while(tcase--)    {     long long l, r, k;     scanf("%lld%lld%lld",&l,&r,&k);     long long cnt=0;     for(i=l;i<=r;i++)     {         //printf("%d",quickpow(i, k));         cnt += divNum(quickpow(i,k))%mod;     }     printf("%lld\n",cnt);    }