LightOJ1007(欧拉函数)

Mathematically Hard

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

分析：题目让求区间[a,b]所有数的欧拉函数值之和。

#include<cstdio>
unsigned long long a[5001000];
void Euler()
{
    for(int i=1;i<5000001;i++) a[i]=i; 
    for(int i=2;i<5000001;i++)
    {
        if(a[i]==i)
        {
            for(int j=i;j<5000001;j+=i)
            a[j]=a[j]/i*(i-1);
        }
    }
}
int main()
{
    int T,cas=1;
    Euler();
    for(int i=2;i<5000001;i++)
    a[i]=a[i]*a[i]+a[i-1];
    scanf("%d",&T);
    while(T--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        printf("Case %d: %llu
",cas++,a[y]-a[x-1]);
    }
    return 0;
}