LightOJ 1007(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
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
题解：直接计算每个数的欧拉值然后平方和计算前缀和再相减就好了
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