hdu6069(大区间素数)

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 3713    Accepted Submission(s): 1366


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<
e2cd
/span>

 

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

题意: 给出 l, r, k．求：(lambda d(i^k))mod998244353，其中 l <= i <= r, d(i) 为 i 的因子个数．

思路：若 x 分解成质因子乘积的形式为 x = p1^a1 * p2^a2 * ... * pn^an，那么 d(x) = (a1 + 1) * (a2 + 1) * ... * (an + 1) .显然 d(x^k) = (a1 * k + 1) * (a2 * k + 1) * ... * (an * k + 1) .

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn=1000100;
const ll mod=998244353;
ll prime[maxn]; //第i个素数
bool is_prime[maxn]; //is_prime[i]为true表示i是素数
ll sum[maxn]; //代表每个数的k次的因子个数
ll a[maxn];
//返回n以内的素数的个数
int sieve(int n)
{
int p = 0;
memset(is_prime,true,sizeof(is_prime));
is_prime[0] = is_prime[1] = false;
for (int i = 2; i <= n; i++){
if (is_prime[i]){
prime[p++] = i;
for (int j = 2 * i; j <= n; j += i)
is_prime[j] = false;
}
}
return p;
}
int main()
{
int p1=sieve(maxn-1);
int t;
ll l,r,k;
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;
}
for(int i=0;prime[i]*prime[i]<=r;i++)
{
ll x=(l/prime[i]+(l%prime[i]?1:0))*prime[i];
for(ll p=x;p<=r;p+=prime[i])
{
ll tmp=0;
while(a[p-l]%prime[i]==0)
{
tmp++;
a[p-l]/=prime[i];
}
sum[p-l]=(sum[p-l]*(tmp*k+1)%mod)%mod;
}
}
ll ans=0;
for(int i=0;i<=r-l;i++)
{
//其含有的质因子中有大于1e6的质数，需要加入计算
if(a[i]!=1)sum[i]=(sum[i]*(k+1))%mod;
ans=(ans+sum[i])%mod;
}
printf("%lld\n",ans);
}
}