LightOJ-1007-Mathematically Hard-欧拉函数打表+前缀和+预处理

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 froma 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

Note

题意：

　　　不是求区间内素数个数的平方和，是求区间内欧拉函数值的平方和

思路：欧拉函数打表（模板）+前缀和+预处理不然TLE

求欧拉函数1-N的打表模板：

 void init()
 {
     memset(ans,,sizeof(ans));//一定要清空
     ans[]=;
     for(int i=; i<=N; i++)
     {
         if(ans[i]==)
         {
             for(int j=i; j<=N; j+=i)
             {
                 if(ans[j]==)
                 {
                     ans[j]=j;
                 }
                 ans[j]=ans[j]/i*(i-);
             }

         }
     }
 }

 #include<stdio.h>
 #include<map>
 #include<string.h>
 #include<iostream>
 typedef long long ll;
 using namespace std;

 const int N=*1e6+;
 unsigned long long ans[N];//unsigned long long输出是%llu，long long会爆，前者20位，后者19位

 void init()
 {
     memset(ans,,sizeof(ans));//一定要清空，不然WA
     ans[]=;
     for(int i=; i<=N; i++)
     {
         if(ans[i]==)
         {
             for(int j=i; j<=N; j+=i)
             {
                 if(ans[j]==)
                 {
                     ans[j]=j;
                 }
                 ans[j]=ans[j]/i*(i-);
             }

         }
     }//欧拉函数打表
     for(int i=; i<=N; i++)
     {
         ans[i]=ans[i-]+ans[i]*ans[i];//计算前缀和
     }
 }
 int main()
 {
     init();
     int tt=,t;
     scanf("%d",&t);
     while(t--)
     {
         int a,b;
         scanf("%d %d",&a,&b);
         printf("Case %d: %llu\n",tt++,ans[b]-ans[a-]);//取区间，左端区间-1，别弄错了
     }

     return ;
 }