hdu6069 Counting Divisors 质因数分解 区间筛

http://acm.hdu.edu.cn/showproblem.php?pid=6069

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2757    Accepted Submission(s): 1020

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

10482302

 

Source

2017 Multi-University Training Contest - Team 4

 

题意：d[i]：表示i的所有因子个数。现在求∑d(i^k)，i∈[l,r]。

题解：首先每个数都可以分解成：i=p1^c1*p2^c2*...*pn^cn，其中p1，p2，...，pn均为质数。那么d[i]就可以表示成(p1+1)*(p2+1)*...*(pn+1)，所以d[i^k]=(p1*k+1)*(p2*k+1)*...*(pn*k+1)。所以本题先对素数打表，但是直接枚举l-r仍然会超时。可以反过来枚举每个质数，用筛法的方式去枚举质数的每个倍数，这样就比原来快多了。

代码：

#include<bits/stdc++.h>#define debug cout<<"aaa"<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 1000000 + 5;const LL mod = 998244353;int primes[N],tot=0,t;LL l,r,k;LL num[N],ans[N];bool isPrime[N];void getPrime()//素数筛 {    mem(isPrime,true);    for(int i=2;i<N;i++)    {        if(isPrime[i])            primes[++tot]=i;        for(int j=1;j<=tot;j++)        {            if(i*primes[j]>=N) break;            isPrime[i*primes[j]]=false;            if(i%primes[j]==0) break;        }    }}int main(){getPrime();scanf("%d",&t);while(t--){scanf("%lld%lld%lld",&l,&r,&k);LL n=(r-l+1);for(int i=0;i<n;i++){//保存每个数字和其贡献值 num[i]=i+l;ans[i]=1;}for(int i=1;(LL)primes[i]*primes[i]<=r;i++){//枚举每个素数 for(LL j=primes[i]*(l/primes[i]);j<=r;j+=primes[i]){//枚举每个素数的倍数 if(j<l){continue;}LL cnt=0;while(num[j-l]%primes[i]==0){cnt++;num[j-l]/=primes[i];}ans[j-l]=(ans[j-l]*(1LL+cnt*k))%mod;}}LL sum=0;for(int i=0;i<n;i++){if(num[i]>1){//表示num[i]本身就是素数 ans[i]=(ans[i]*(1LL+k))%mod;}sum=(sum+ans[i])%mod;}printf("%lld\n",sum);}return 0;}