2017 Multi-University Training Contest

题目传送门
Counting Divisors

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1035 Accepted Submission(s): 364

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
题解：
根据约数个数定理：n=p1^a1×p2^a2×p3^a3*…*pk^ak,n的约数的个数就是(a1+1)(a2+1)(a3+1)…(ak+1).
若i=p1^a1×p2^a2×p3^a3*…pk^ak,则i^K=p1^(a1*K)×p2^(a2*K)×p3^(a3*K)…*pk^(ak*K),i^K的约数的个数就是(a1*K+1)(a2*K+1)(a3*K+1)…(ak*K+1)
但是题目重点转换为L~R的因数个数和，也就是重点转换为L~R的质因数分解，两次塞选，第一次塞出1e6以内的所有质数，第二次枚举区间[L,R]中之前塞选得到质数的倍数，剩下的就是[L,R]的质数，用cnt数组记录它们因数的个数，因为L和R比较大，所以再开一个数组存储它们的值，所以cnt[i]就可以相当于cnt[i+L]。
要用long long，不然中间会爆int。
AC代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>typedef long long ll;using namespace std;const int maxn=1e6+10;const int mod=998244353;int tot,t;ll l,r,k,ans,cnt[maxn],q[maxn],prime[maxn];bool vis[maxn];void init(){    for(int i=2;i<maxn;i++){        if(!vis[i])            prime[tot++]=i;        for(int j=0;j<tot&&i*prime[j]<maxn;j++)        {            vis[i*prime[j]]=1;            if(!(i%prime[j]))            break;        }    }}//快速素数筛 int main(){    scanf("%d",&t);    init();    while(t--)    {        scanf("%lld%lld%lld",&l,&r,&k);        ans=0;        if(l==1)//1它只有一个因数（自己）比较独特，特判一下         ans++,l++;/*所以ans+=1,l->2（一定要变成2，不然wa,不过也可以改后面的代码，也可ac,比较麻烦）*/        for(ll i=0;i<=r-l;i++)         cnt[i]=1,q[i]=l+i;        for(ll i=0;prime[i]*prime[i]<=r;i++)        {            ll j=l/prime[i]+(l%prime[i]!=0);            for(j=j*prime[i];j<=r;j+=prime[i])            {                ll tmp=0;                while(q[j-l]%prime[i]==0)                q[j-l]/=prime[i],tmp++;                cnt[j-l]*=(tmp*k+1)%mod,cnt[j-l]%=mod;            }        }//将l-r的非质数进行分解，分解成质因数         for(ll i=0;i<=r-l;i++)        {            if(q[i]!=1)             ans+=((k+1)*cnt[i])%mod;            /*(1）i为素数时            (2）i为非素数,但在上面分解成质因数时，q[i]!=1,所以最后q[i]一定是一个素数所以结合（1）（2）最后是乘（k*1+1）=(k+1)*/             else                    ans+=cnt[i];            ans%=mod;        }        printf("%lld\n",ans);    }}