HDOJ 6069 素数筛法（数学）

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 3041    Accepted Submission(s): 1130

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

3
1 5 1
1 10 2
1 100 3

 

Sample Output

10
48
2302

 

Source

2017 Multi-University Training Contest - Team 4

 

思路：先将1000005以内的素数打表(素数筛法O(n))，然后将每个数拆成S=p1^a1*p2*^a2...*pn^an 的形式则 d(S)=(1+a1)*(1+a2)....*(1+an);

实现方法：用num[] 存下当前拆分后剩下的数字，用ans[i]存下第i个数字此时的已拆分的累乘结果，将素数从小到大扫一遍，对[l,r]每个数字进行拆分，逐一更新ans[i]的值

代码：

#include<bits/stdc++.h>

typedef long long ll;
using  namespace std;

;
const int inf=1e9;
;

ll l,r,k;
bool v[N];
ll ans[N];
ll num[N];
int pri[N];
int pcnt;
void init()
{
    pcnt = ;//素数筛
    ; i <= N; i++)
    {
        if(!v[i])
            pri[pcnt++] = i;
        ; j < pcnt && pri[j] <= N/i; j++)
        {
            v[i*pri[j]] = ;
            ) break;
        }
    }
}
void f()
{
    for(ll i=l;i<=r;i++)//初始化记录数组
        num[i-l]=i,ans[i-l]=;
}
int main()
{
    int t;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&l,&r,&k);
        f();
        ;i<pcnt&&pri[i]<=;i++){//限制质因子大小
            ll p=pri[i];
            ll d=l/p+(l%p>);
            ) d=;
            for(ll j=d*p;j<=r;j+=p){
                ll cnt=;
                ){
                    num[j-l]/=p;
                    cnt++;
                }
                ans[j-l]=(ans[j-l]%mod)*((+k*cnt)%mod)%mod;
            }
        }
        ll sum=;
        for(ll i=l;i<=r;i++){
            )//还存在一个大于1000000的因子，再乘上k+1
                sum=(sum+ans[i-l]*(k+))%mod;
            else
                sum=(sum+ans[i-l])%mod;
        }
        printf("%lld\n",sum);
    }

}