多校4 HDU6069 - Counting Divisors 素数、分解质因数

原题链接：

HDU-6069

Counting Divisors

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

Total Submission(s): 2400 Accepted Submission(s): 870

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 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

3

1 5 1

1 10 2

1 100 3

Sample Output

10

48

2302

思路：

这道题一直 T ，赛后看题解也算是有所收获，在分解质因数的地方学会了一点优化。



具体思路：

生成素数筛，简单地利用质因数的计算方法即可，关键在于分解质因数的时候做一点优化。

算质数的倍数来分解质因数。

见代码。

PS.预处理的时候不要舍本逐末,瞎搞反而增加时间….

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mem(s,t) memset(s,t,sizeof(s))
#define D(v) cout<<#v<<" "<<v<<endl
#define inf 0x3f3f3f3f
//#define LOCAL
const ll MAXM =80000;//1e6范围内质数数量
const ll mod=998244353;
const ll MAXN = 1e6+10;
ll l,r,k,num[MAXN],cnt[MAXN];
bool isprime[MAXN];
ll tot =0;
ll prime[MAXM];
void getprime() {
mem(isprime,true);
isprime[1]=false;
tot=0;
for(ll i=2; i<MAXN; i++) {
if(isprime[i]) prime[++tot]=i;
for(ll j=1;j<=tot && i*prime[j]<MAXN;j++) {
isprime[i*prime[j]]=false;
if(i%prime[j]==0)break;
}
}
}
int main() {
getprime();
int t;
scanf("%d",&t);
while(t--){
scanf("%lld%lld%lld",&l,&r,&k);
ll n=r-l;
for(ll i=0;i<=n;i++){
num[i]=l+i;
cnt[i]=1;
}
for(ll i=1;i<=tot;i++){
if(prime[i]*prime[i]>r) break;
for(ll j=l/prime[i]*prime[i];j<=r;j+=prime[i]) if(j>=l){
ll o=0;
while(num[j-l]%prime[i]==0){
o++;
num[j-l]/=prime[i];
}
cnt[j-l]=cnt[j-l]*(o*k%mod+1)%mod;
}
}
ll ans=0;
for(ll i=0;i<=n;i++){
if(num[i]>1) cnt[i]=cnt[i]*(k+1)%mod;
//D(cnt[i]);
ans=(ans+cnt[i])%mod;
}
printf("%lld\n",ans);
}
return 0;
}

附 标程：

#include<cstdio>
typedef long long ll;
const int N=1000010,P=998244353;
int Case,i,j,k,p[N/10],tot,g
,ans;ll n,l,r,f
;bool v
;
inline void work(ll p){
for(ll i=l/p*p;i<=r;i+=p)if(i>=l){
int o=0;
while(f[i-l]%p==0)f[i-l]/=p,o++;
g[i-l]=1LL*g[i-l]*(o*k+1)%P;
}
}
int main(){
for(i=2;i<N;i++){
if(!v[i])p[tot++]=i;
for(j=0;j<tot&&i*p[j]<N;j++){
v[i*p[j]]=1;
if(i%p[j]==0)break;
}
}
scanf("%d",&Case);
while(Case--){
scanf("%lld%lld%d",&l,&r,&k);
n=r-l;
for(i=0;i<=n;i++)f[i]=i+l,g[i]=1;
for(i=0;i<tot;i++){
if(1LL*p[i]*p[i]>r)break;
work(p[i]);
}
for(ans=i=0;i<=n;i++){
if(f[i]>1)g[i]=1LL*g[i]*(k+1)%P;
ans=(ans+g[i])%P;
}
printf("%d\n",ans);
}
return 0;
}