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  • HDU 6069 Counting Divisors

    Counting Divisors

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
    Total Submission(s): 1604    Accepted Submission(s): 592


    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(1T15), denoting the number of test cases.

    In each test case, there are 3 integers l,r,k(1lr1012,rl106,1k107).
     
    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
     
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    /*
    * @Author: Lyucheng
    * @Date:   2017-08-03 13:13:45
    * @Last Modified by:   Lyucheng
    * @Last Modified time: 2017-08-04 11:25:19
    */
    /*
     题意:给你一个区间[l,r]让你求区间内每个数的k次方因子数的总和
    
     思路:比赛的时候想出来 i^k的因子是 (n1*k+1)*(n2*k+1)*...*(np*k+1),但是没想出来怎么优化,素数枚举
        很烦,四场比赛每次差一点,比赛的时候想的是枚举[l,r]之间的数,优化到8300ms,实在没法优化了,应
        该反过来想,枚举从[l,r]的素因子,因为i素因子的个数远小于i,大多数在sqrt i内,最多只有一个在sqrt i
        之外。
    */
    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <time.h>
    
    #define LL long long
    #define mod 998244353
    #define MAXN 1000005
    
    using namespace std;
    
    int t;
    LL l,r,k;
    LL res;
    LL d[MAXN];
    LL pos[MAXN];
    bool prime[MAXN];
    LL p[MAXN];
    int tol;
    void init(){
        tol=0;
        for(int i=0;i<MAXN;i++){
            prime[i]=true;
        }
        prime[0]=prime[1]=false;
        for(int i=2;i<MAXN;i++){
            if(prime[i]) 
                p[tol++]=(LL)i;
            for(int j=0;j<tol&&i*p[j]<MAXN;j++){
                prime[i*p[j]]=false;
                if((LL)i%p[j]==0) break;
            }
        }
    }
    
    int main(){
        // freopen("in.txt", "r", stdin);
        // freopen("out.txt", "w", stdout);
        init();
        scanf("%d",&t);
        while(t--){
            res=0;
            scanf("%I64d%I64d%I64d",&l,&r,&k);
            for(LL i=l;i<=r;i++){
                d[(int)(i-l)]=1;
                pos[(int)(i-l)]=i;
            }
            for(int i=0;i<tol;i++){//枚举所有的素数
                if(p[i]<=0) continue;
                LL cnt=(l+p[i]-1)/p[i]*p[i];//找出[l,r]区间内的第一个p[i]的倍数
                if(cnt-l<0||cnt-l>r-l) continue;
                for(int j=cnt-l;j<=r-l;j+=p[i]){
                    LL cur=0;
                    while(pos[j]&&pos[j]%p[i]==0){
                        cur++;
                        pos[j]/=p[i];
                    }
                    d[j]*=(cur*k+1);
                    d[j]%=mod;
                }
            }
            for(int i=0;i<=r-l;i++){
                if(pos[i]==1)
                    res+=d[i];
                else
                    res+=d[i]*(k+1);
                res%=mod;
            }
            printf("%I64d
    ",res);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wuwangchuxin0924/p/7284093.html
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