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

31 5 11 10 21 100 3

 

Sample Output

10482302

 

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。