HDU 6069 Counting Divisors(2017 Multi-University Training Contest - Team 4 )

  

 

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：就是那个公式

 

感觉还是题解讲的清楚

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype> 
#include<cmath>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
#include<string> 
#define ll long long
#define eps 1e-10
#define LL unsigned long long
using namespace std; 
const int maxn=1000000+100;
const int mod=998244353;
int vis[maxn];
int prim[maxn];
ll a[maxn];
ll b[maxn];
int len;
void get_prim() 
{
    memset(vis,0,sizeof(vis));
    int m=sqrt(maxn+0.5);
    for(int i=2;i<=m;i++)
    if(vis[i]==0)
    for(int j=i*i;j<=maxn;j+=i)
    vis[j]=1;
    len=0;
    for(int i=2;i<=maxn;i++)
    if(vis[i]==0)
    prim[len++]=i;
}
int main()
{
    int t;
    ll l,r,k;
    get_prim();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&l,&r,&k);
        for(int i=0;i<maxn;i++)
        {
          a[i]=1;
          b[i]=i+l;
        }
          for(int i=0;i<len;i++)
          {
              ll cnt=l;
            if(l%prim[i])cnt=l+prim[i]-l%prim[i];
              for(ll j=cnt;j<=r;j+=prim[i])
              {
                  int count=0;
                  while(b[j-l]%prim[i]==0)
                  {
                      count++;
                      b[j-l]/=prim[i];
                }
                a[j-l]=((k*count+1)%mod*a[j-l])%mod;
              }
          }
          ll ans=0;
          for(int i=0;i<=r-l;i++)
          if(b[i]>1)
          a[i]=(a[i]*(k+1))%mod;
          for(int i=0;i<=r-l;i++)
          {
              ans=(ans+a[i])%mod;
          }
          printf("%lld\n",ans);
    }
    return 0;
}