LightOJ1007 Mathematically Hard 欧拉函数+前缀和

Description

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

3

6 6

8 8

2 20

Sample Output

Case 1: 4

Case 2: 16

Case 3: 1237

Hint

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

#include <iostream>#include <cstdio>#include <map>#include <set>#include <vector>#include <queue>#include <stack>#include <cmath>#include <algorithm>#include <cstring>#include <string>using namespace std;#define INF 0x3f3f3f3ftypedef unsigned long long LL;#define maxn 5000000LL euler[maxn+5];void euler_init(){    for(int i=0;i<=maxn;i++){        euler[i]=i;    }    for(int i=2;i<=maxn;i++){        if(euler[i]==i){            for(int j=i;j<=maxn;j+=i){                euler[j]=euler[j]/i*(i-1);            }        }    }    for(int i=2;i<=maxn;i++){        euler[i]=euler[i-1]+euler[i]*euler[i];    }}int main(){std :: ios :: sync_with_stdio (false);    int t,a,b,ca=1;    scanf("%d",&t);    euler_init();    while(t--){        scanf("%d%d",&a,&b);        printf("Case %d: %llu\n",ca++,euler[b]-euler[a-1]);    }    return 0;}