1007 - Mathematically Hard

1007 - Mathematically Hard

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 64 MB

Mathematically some problems look hard. But with the help of the computer, some problems can be easily solvable.

In this problem, you will be given two integers a and b. You have to find the summation of the scores of the numbers from a to b (inclusive). The score of a number is defined as the following function.

score (x) = n², where n is the number of relatively prime numbers with x, which are smaller than x

For example,

For 6, the relatively prime numbers with 6 are 1 and 5. So, score (6) = 2² = 4.

For 8, the relatively prime numbers with 8 are 1, 3, 5 and 7. So, score (8) = 4² = 16.

Now you have to solve this task.

Input

Input starts with an integer T (≤ 10⁵), denoting the number of test cases.

Each case will contain two integers a and b (2 ≤ a ≤ b ≤ 5 * 10⁶).

Output

For each case, print the case number and the summation of all the scores from a to b.

Sample Input	Output for Sample Input
3 6 6 8 8 2 20	Case 1: 4 Case 2: 16 Case 3: 1237

Note

Euler's totient function  applied to a positive integer n is defined to be the number of positive integers less than or equal to n that are relatively prime to n.  is read "phi of n."

Given the general prime factorization of , one can compute  using the formula

思路：就是求欧拉函数，先素数打表，再欧拉函数打表，最后求下前缀和；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<queue>
using namespace std;
bool pr[5*1000006]={0};
int prime[5*100006];
unsigned long long  oula[5*1000006];
typedef long long LL;
int main(void)
{
  LL i,j,p,q;
  int s,k;
  scanf("%d",&k);pr[0]=true;
  pr[1]=true;
  for(i=0;i<5*1000006;i++)
    oula[i]=i;
  for(i=2;i<10000;i++)
  {
      if(!pr[i])
      {
          for(j=i;i*j<=5*1000000;j++)
          {
              pr[i*j]=true;
          }
      }
  }int cnt=0;
  for(i=2;i<5*1000005;i++)
  {
      if(pr[i]==false)
        prime[cnt++]=i;
  }
  for(i=0;i<cnt;i++)
  {
      for(j=1;prime[i]*j<=5*1000000;j++)
      {
          oula[prime[i]*j]=oula[prime[i]*j]/prime[i]*(prime[i]-1);
      }
  }
  for(i=2;i<=1000000*5;i++)
  {
      oula[i]=oula[i-1]+oula[i]*oula[i];
  }
  for(s=1;s<=k;s++)
  {
      scanf("%lld %lld",&p,&q);printf("Case %d: ",s);
      printf("%llu\n",oula[q]-oula[p-1]);
  }
  return 0;
}