Hdu6069 Counting Divisors（2017多校第4场）

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Counting Divisors

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1105 Accepted Submission(s): 393

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 :

(\sum i = l r d (i k)) mod 998244353

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

Sample Output

Source

2017 Multi-University Training Contest - Team 4

Recommend

liuyiding

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题目的意思是求一个区间的每个数的因子数之和

思路：先去打一个1到1e6的素数表，然后去枚举每个素数在区间内的倍数，可以跳着枚举，计算出每个数对应的因子个数，对于每个数的因子个数就等于枚举的因子个数*k+1累乘起来，注意剩下的大素数的判断

比赛是也是素数打表枚举的，鬼知道怎么想的去枚举区间的每个数去求因子，妥妥TLE，也有过正解的想法，想过没想太透彻，感觉这种题值得好好反思警戒自己

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const LL mod = 998244353;

bool vis[1000009];
LL prime[1000009], sum[1000009], a[1000009];
LL l, r, k;
int cnt;

void init()
{
    memset(vis, true, sizeof vis);
    vis[0] = vis[1] = false;
    cnt = 0;
    for (int i = 2; i < 1000009; i++)
    {
        if (!vis[i]) continue;
        prime[cnt++] = i;
        for (int j = i * 2; j < 1000009; j += i) vis[j] = false;
    }
}

int main()
{
    init();
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld%lld%lld", &l, &r, &k);
        for (int i = 0; i <= r - l; i++) sum[i] = 1, a[i] = i + l;
        LL ans = 0;
        for (int i = 0; i < cnt; i++)
        {
            LL p = (l / prime[i] + (l%prime[i] ? 1 : 0))*prime[i];
            for (LL j = p; j <= r; j += prime[i])
            {
                int res = 0;
                while (a[j - l] % prime[i] == 0) res++, a[j - l] /= prime[i];
                sum[j - l] = sum[j - l] * ((1LL * res*k + 1) % mod) % mod;
            }
        }
        for (int i = 0; i <= r - l; i++)
        {
            if (a[i] != 1) sum[i] = sum[i] * (k + 1) % mod;
            ans = (ans + sum[i]) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}