LightOJ1007 - Mathematically Hard

1007 - Mathematically Hard

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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 anumber 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

题目大意：

求a~b之间的欧拉函数的平方的和；

解题思路：

其实如果你知道欧拉函数的话这道题的解题思路就比较简单了，但是由于数据比较大，所以应该用unsigned long long 类型的数据来保存。至于什么是欧拉函数这里就不在讲解了，百度百科或者维基百科讲的都比较的详细。

#include<stdio.h>#include<string.h>#include<malloc.h>#define N 5000009typedef unsigned long long  ULL;ULL *phi=(ULL *)malloc(N*sizeof(ULL));void getPhi(){    memset(phi,0,sizeof(phi));    int i,j;    phi[1]=1;  for(i=2;i<=N;i++)    if(!phi[i])       for(j=i;j<=N;j+=i){            if(!phi[j])                 phi[j]=j;                 phi[j]=phi[j]/i*(i-1);       }}void initi(){   int i;   ULL temp;   phi[0]=0;   for(i=1;i<N;i++){       temp+=phi[i]*phi[i];       phi[i]=temp;   }}int main(){    getPhi();    initi();    int t,i;    scanf("%d",&t);    int a,b;    i=1;    while(t--){            scanf("%d%d",&a,&b);            printf("Case %d: %llu\n" , i , phi[b] - phi[a-1]) ;            ++i;    }    return 0;}