Hdu6069 Counting Divisors（2017多校第4场）

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 :

(∑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

 

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 longconst 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;}