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,0,sizeof(ans));//一定要清空
    ans[1]=1;
    for(int i=2; i<=N; i++)
    {
        if(ans[i]==0)
        {
            for(int j=i; j<=N; j+=i)
            {
                if(ans[j]==0)
                {
                    ans[j]=j;
                }
                ans[j]=ans[j]/i*(i-1);
            }

        }
    }
}

 

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

const int N=5*1e6+10;
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,0,sizeof(ans));//一定要清空，不然WA
    ans[1]=1;
    for(int i=2; i<=N; i++)
    {
        if(ans[i]==0)
        {
            for(int j=i; j<=N; j+=i)
            {
                if(ans[j]==0)
                {
                    ans[j]=j;
                }
                ans[j]=ans[j]/i*(i-1);
            }

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

    return 0;
}