hdu 6069 Counting Divisors（2017多校第四场）

题目链接：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): 3282    Accepted Submission(s): 1221

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

 

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解析：n = k1^x1*k2^x2*........kn^xn， （k1, k2, k3.......kn是质数, x1, x2,z3.......xn是幂），那么一个数的因子数等于 ans = (x1+1)*(x2+1)*(x3+1)*......*(xn+1)。知道这样还不够，这样暴力写会TLE，咱还需要用到区间筛选，对，就是这样，打素数表时10^5会WA，需要打10^6的表，WA的心疼，，

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1000006;const LL p = 998244353;int pri[N], ispri[N], len;LL a[1000009], num[1000009];void init(){    memset(ispri, 0, sizeof(ispri));    len = 0;    for(int i = 2; i <= 1000000; i++)    {        if(ispri[i]) continue;        pri[len++] = i;        for(int j = i*2; j <= 1000000; j += i) ispri[j] = 1;    }}int main(){    init();    LL l, r, k;    int t;    scanf("%d", &t);    while(t--)    {        scanf("%lld%lld%lld", &l, &r, &k);        for(int i = 0; i <= r - l; i++) a[i] = l+i, num[i] = 1;        for(int i = 0; i < len; i++)        {            LL s = l/pri[i]*pri[i];            if(s != l) s += pri[i];            for(LL j = s; j <= r; j += pri[i])            {                LL w = 0;                while(a[j-l] % pri[i] == 0) w++, a[j-l] /= pri[i];                num[j-l] = num[j-l] * ((1LL*w*k+1) % p) % p;            }        }        LL ans = 0;        for(int i = 0; i <= r-l; i++) if(a[i] != 1) num[i] = (num[i] * (k + 1)) % p;        for(int i = 0; i <= r-l; i++) ans = (ans + num[i]) % p;        printf("%lld\n", ans);    }    return 0;}