2017 Multi-University Training Contest

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2340    Accepted Submission(s): 841

Problem Description

In mathematics, the functiond (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 integerT(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

 

题目大意：

求给定的函数值，其中 d (i)：i 的因子个数

解题思路：

设有 n=pa11∗pa22∗...∗pakk，其中n 的因子个数为：(a1+1)∗(a2+1)∗...∗(ak+1)

首先预处理出1-10^6里面所有的素数，然后枚举这些素数，在枚举[L,R]区间中这些素数的倍数，然后根据这些倍数进行素因子分解，其实就是统计[L,R]区间中有多少个枚举的素数，然后乘以K， 在加 1 计算即可。在枚举 [L,R]区间值的时候，我们需要先把[L,R] 区间中的数用数组保存下来，然后通过数组进行素因子分解。

#include <bits/stdc++.h>using namespace std;typedef long long LL;const LL MOD = 998244353;const LL MAXN = 1e6 + 5;int prime[MAXN], cnt;LL p[MAXN];void isprime(){    memset(prime, 0, sizeof(prime));    cnt = 0;    prime[1] = 1;    for (LL i = 2; i < MAXN; i++)    {        if (!prime[i])        {            p[cnt++] = i;            for (LL j = i * i; j < MAXN; j += i)            {                prime[j] = 1;            }        }    }}LL f[MAXN], num[MAXN];int main(){    isprime();    int T;    scanf("%d", &T);    while (T--)    {        LL L, R, K;        scanf("%lld%lld%lld", &L, &R, &K);        LL ans = 0;        if (L == 1)        {            ans = 1, L++;        }        int t = R - L;        for (int i = 0; i <= t; i++)        {            f[i] = i + L, num[i] = 1;        }        for (int i = 0; i < cnt && p[i]*p[i] <= R; i++)        {            LL tmp = L;            if (L % p[i])            {                tmp = (L / p[i] + 1) * p[i];            }            for (LL j = tmp; j <= R; j += p[i])            {                LL cnt = 0;                while (f[j - L] % p[i] == 0)                {                    cnt++;                    f[j - L] /= p[i];                }                num[j - L] = num[j - L] * (cnt * K + 1) % MOD;            }        }        for (int i = 0; i <= t; i++)        {            if (f[i] == 1)            {                ans = (ans + num[i]) % MOD;            }            else            {                ans = (ans + num[i] * (K + 1)) % MOD;            }        }        printf("%lld\n", ans);    }    return 0;}