HDU 6069-Counting Divisors（多校训练第四场->区间质因数个数）

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1952    Accepted Submission(s): 710

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

 

Recommend

题意：见题面吧。。。

题解：比赛的时候简直智障，感觉暴力必定过不了我就没太敢下手，后来现场搞一发Pollard_rho算法，改了将近一个多小时，还是TLE，这么难！！！果断放弃了。（还是太菜），今天补了这道题，发现和暴力的唯一区别就是区间枚举质因数的时候不是++，而是+p，woc，为什么比赛的时候没想到，就一个字母的事情，真的是菜菜菜菜菜菜啊。。。

我当时是怎么想的，直接在区间里枚举当前质因数的倍数就好了233333.。。。

#include<map>              #include<stack>              #include<queue>            #include<vector>              #include<math.h>        #include<time.h>      #include<stdio.h>            #include<iostream>          #include<string.h>              #include<stdlib.h>              #include<algorithm>              using namespace std;              typedef long long  ll;              #define inf 2147483647             #define mod 998244353          #define maxn  1000005          #define lowbit(x) (x&-x)              #define eps 1e-10      ll a[maxn]={1,1},b[maxn],c[maxn],d[maxn];int main(void){ll T,l,r,i,j,ans,cnt=0,tmp,k,res;for(i=2;i<=maxn;i++){if(a[i])continue;b[++cnt]=i;for(j=i*i;j<=maxn;j+=i)a[j]=1;}scanf("%lld",&T);while(T--){ans=0;scanf("%lld%lld%lld",&l,&r,&k);for(i=l;i<=r;i++)c[i-l]=i,d[i-l]=1;for(i=1;b[i]*b[i]<=r && i<=cnt;i++){tmp=l-l%b[i];if(tmp<l)tmp+=b[i];for(j=tmp;j<=r;j+=b[i]){res=0;if(c[j-l]%b[i]==0){while(c[j-l]%b[i]==0)c[j-l]/=b[i],res++;d[j-l]=(d[j-l]*(1+k*res))%mod;}}}for(i=l;i<=r;i++)if(c[i-l]>1)d[i-l]=(d[i-l]*(1+k))%mod;for(i=l;i<=r;i++)ans=(ans+d[i-l])%mod;printf("%lld\n",ans);}return 0;}