light oj 1007 Mathematically Hard （欧拉函数）_for m n euler function m is defined to be the numb-优快云博客

本文链接：https://blog.youkuaiyun.com/zs120197/article/details/77452955

本文介绍了一道涉及欧拉函数的数学编程题，题目要求计算特定范围内所有整数的欧拉函数值平方和。文章提供了详细的解题思路，包括使用欧拉函数计算互质数个数的方法，并给出了优化过的C++代码实现。

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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

6 6

8 8

2 20

Sample Output

Case 1: 4

Case 2: 16

Case 3: 1237

Hint

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

题意：假设所有小于等于x的数中，与x互质的数的个数为n，则sorce(x)=n的平方。给你a和b，求从a到b所有的sorce(i)的和。

其实就是一道简单的欧拉函数的模板题，但是坑点实在太多。

坑点：

1、会爆long long导致wa，应该定义为unsigned long long。

2、不能开两个数组，不然会爆内存。求欧拉函数值的数组和前缀和的数组可以用一个数组表示。

3、不能用cout输出，不然会超时，应该用printf。

#include <bits/stdc++.h>
#define N 5000000+10
using namespace std;
typedef unsigned long long ll;
ll Euler[N];

void  init()
{
    Euler[1]=1;
    for(int i=2; i<N; i++)
        Euler[i]=i;
    for(int i=2; i<N; i++)
        if(Euler[i]==i)
            for(int j=i; j<N; j+=i)
                Euler[j]=Euler[j]/i*(i-1);
}
void solve()
{
    for(int i=2; i<N; i++)
        Euler[i]=Euler[i-1]+Euler[i]*Euler[i];
}

int main()
{
    init();
    solve();
    int t,a,b;
    cin>>t;
    for(int cas=1; cas<=t; cas++)
    {
        scanf("%d%d",&a,&b);
        printf("Case %d: %llu\n",cas,Euler[b]-Euler[a-1]);
    }
    return 0;
}