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=+;
 const int mod=;
 int vis[maxn];
 int prim[maxn];
 ll a[maxn];
 ll b[maxn];
 int len;
 void get_prim()
 {
     memset(vis,,sizeof(vis));
     int m=sqrt(maxn+0.5);
     for(int i=;i<=m;i++)
     if(vis[i]==)
     for(int j=i*i;j<=maxn;j+=i)
     vis[j]=;
     len=;
     for(int i=;i<=maxn;i++)
     if(vis[i]==)
     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=;i<maxn;i++)
         {
           a[i]=;
           b[i]=i+l;
         }
           for(int i=;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=;
                   while(b[j-l]%prim[i]==)
                   {
                       count++;
                       b[j-l]/=prim[i];
                 }
                 a[j-l]=((k*count+)%mod*a[j-l])%mod;
               }
           }
           ll ans=;
           for(int i=;i<=r-l;i++)
           if(b[i]>)
           a[i]=(a[i]*(k+))%mod;
           for(int i=;i<=r-l;i++)
           {
               ans=(ans+a[i])%mod;
           }
           printf("%lld\n",ans);
     }
     return ;
 }