Mathematically Hard（欧拉函数打表+前缀和）

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

题意·：给出两个数，确定一个范围，输出范围内数字的欧拉函数的平方的和。
欧拉函数：在数论，对正整数n，欧拉函数是小于或等于n的正整数中与n互质的数的数目；

欧拉函数打表：（对于没理解欧拉函数推导的我来说只能先记代码）

    for(i=0;i<=5010000;i++)
	a[i]=0;
	a[1]=1;
	for(i=2;i<=5010000;i++)
	{
		if(!a[i])
		{
			for(j=i;j<=5010000;j+=i)
			{
				if(!a[j])
				a[j]=j;
				a[j]=a[j]/i*(i-1);
			}
		}
	}

需要特别注意的是，打表的数组要用unsigned long long定义，直接用long long会爆，前者是20位，后者是19位。

代码：

#include"stdio.h"
unsigned long long a[5100000];//注意类型
void init()
{
	int i,j;
	for(i=0;i<=5010000;i++)
	a[i]=0;
	a[1]=1;
	for(i=2;i<=5010000;i++)
	{
		if(!a[i])
		{
			for(j=i;j<=5010000;j+=i)
			{
				if(!a[j])
				a[j]=j;
				a[j]=a[j]/i*(i-1);
			}
		}
	}
	for(i=2;i<=5010000;i++)//前缀和，注意是平方的和
	a[i]=a[i]*a[i]+a[i-1];
}
int main()
{
	init();
	int t,k=1;
	scanf("%d",&t);
	while(t--)
	{
		int m,n;
		scanf("%d %d",&m,&n);
		printf("Case %d: %llu\n",k++,a[n]-a[m-1]);
	}
	return 0;
}