多校

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1286    Accepted Submission(s): 454

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

自己写的代码：

#include<cstdio>
typedef long long ll;
const int N=1000010,P=998244353;
int Case,i,j,k,p[N/10],tot,g[N],ans;
ll n,l,r,f[N];
bool v[N];
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;
        }
    }
//    素数筛法
    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;
}

显然服务器只要稍微一不高兴，就给你卡了。

陈老师的代码：

#include<cstdio>
typedef long long ll;
const int N=1000010,P=998244353;
int Case,i,j,k,p[N/10],tot,g[N],ans;
ll n,l,r,f[N];
bool v[N];
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;
        }
    }
//    素数筛法
    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;
}

无论怎么卡也可以过。这个就是实现问题吧。