hdu 6069Counting Divisors 数学

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

31 5 11 10 21 100 3

 

Sample Output

10482302

 

Source

2017 Multi-University Training Contest - Team 4

import java.io.BufferedReader;import java.io.InputStream;import java.io.InputStreamReader;import java.io.PrintWriter;import java.math.BigInteger;import java.util.Arrays;import java.util.StringTokenizer;public class Main {public static void main(String[] args) {new Task().solve();}}class Task {InputReader in = new InputReader(System.in) ;PrintWriter out = new PrintWriter(System.out) ;final int N = 1000000 ;    long[] prime = new long[N+1] ;    int primeCnt ;    boolean[] vis = new boolean[N+1] ;    {    Arrays.fill(vis, false) ;    primeCnt = 0 ;     for(int i = 2 ; i <= N ; i++){    if(! vis[i]){    prime[primeCnt++] = (long)i ;    }    for(int j = 0 ; j < primeCnt && prime[j] * i <= N ; j++){    vis[(int)(i*prime[j])] = true ;      if(i % prime[j] == 0){    break ;    }    }    }    }        final long Mod = 998244353L ;        long calc(long L , long R , long k){          int len = (int)(R - L + 1) ;        long[] num = new long[len] ;        long[] sum = new long[len] ;        for(int i = 0 ; i < len ; i++){        num[i] = L + i ;        sum[i] = 1 ;        }        for(int i = 0 ; i < primeCnt && prime[i]*prime[i] <= R ; i++){              long start = L/prime[i]*prime[i] ;              if(start < L)                  start += prime[i] ;              for(long j = start ; j <= R ; j += prime[i]){                  int idx = (int)(j - L) ;                 if(num[idx] % prime[i] == 0){                long t = 0 ;                while(num[idx] % prime[i] == 0){                num[idx] /= prime[i] ;                t++ ;                }                sum[idx] *=  (t*k + 1) ;                sum[idx] %= Mod ;                }            }        }          long reslut = 0 ;        for(int i = 0 ; i <= R-L ; i++){          if(num[i] != 1){        sum[i] *= (k+1) ;        sum[i] %= Mod ;        }        reslut += sum[i] ;        reslut %= Mod ;        }        return reslut ;      }  void solve(){int t = in.nextInt() ;while(t-- > 0){out.println(calc(in.nextLong(), in.nextLong(), in.nextLong())) ;} out.flush() ;}}class InputReader {        public BufferedReader reader;        public StringTokenizer tokenizer;            public InputReader(InputStream stream) {            reader = new BufferedReader(new InputStreamReader(stream), 32768);            tokenizer = new StringTokenizer("");        }            private void eat(String s) {            tokenizer = new StringTokenizer(s);        }            public String nextLine() {             try {                return reader.readLine();            } catch (Exception e) {                return null;            }        }            public boolean hasNext() {            while (!tokenizer.hasMoreTokens()) {                String s = nextLine();                if (s == null)                    return false;                eat(s);            }            return true;        }            public String next() {            hasNext();            return tokenizer.nextToken();        }            public int nextInt() {            return Integer.parseInt(next());        }            public int[] nextInts(int n) {            int[] nums = new int[n];            for (int i = 0; i < n; i++) {                nums[i] = nextInt();            }            return nums;        }            public long nextLong() {            return Long.parseLong(next());        }            public double nextDouble() {            return Double.parseDouble(next());        }            public BigInteger nextBigInteger() {            return new BigInteger(next());        }        }