hdu 6069 Counting Divisors 筛法

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

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

官方题解：

　　唯一分解定理

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stdlib.h>
#include<time.h>
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define bug(x)  cout<<"bug"<<x<<endl;

const int N=1e5+,M=1e6+,inf=;
const LL INF=1e18+,mod=;

const int MAXN=;
int prime[MAXN];//保存素数
bool vis[MAXN];//初始化
int Prime(int n)
{
    int cnt=;
    memset(vis,,sizeof(vis));
    for(int i=; i<n; i++)
    {
        if(!vis[i])
            prime[cnt++]=i;
        for(int j=; j<cnt&&i*prime[j]<n; j++)
        {
            vis[i*prime[j]]=;
            if(i%prime[j]==)
                break;
        }
    }
    return cnt;
}

LL num[M],ans[M];

int main()
{
    int cnt=Prime();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        for(int i=; i<; i++)
            num[i]=,ans[i]=;
        LL l,r,K;
        scanf("%lld%lld%lld",&l,&r,&K);
        for(int q=; q<cnt; q++)
        {
            int i=prime[q];
            LL L=(l%i==?l:(l/i+)*i);
            for(LL j=L; j<=r; j+=i)
            {
                LL temp=j,base=;
                int x=;
                while(temp%i==)
                {
                    x++;
                    temp/=i;
                    base*=i;
                }
                ans[j-l+]*=(1LL*K*x+)%mod;
                ans[j-l+]%=mod;
                num[j-l+]*=base;
            }
        }
        for(LL i=l; i<=r; i++)
        {
            if(num[i-l+]!=i)
                ans[i-l+]*=K+,ans[i-l+]%=mod;
        }
        LL out=;
        for(LL i=l;i<=r;i++)
            out+=ans[i-l+],out%=mod;
        printf("%lld\n",out);
    }
    return ;
}