HDU 2824 The Euler function

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最新推荐文章于 2021-02-21 18:02:03 发布

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本文介绍了一种计算特定区间内欧拉函数之和的方法。通过预先计算并存储每个数的欧拉函数值，实现了快速查询区间[a,b]内的欧拉函数和。此算法适用于数论研究及竞赛编程。

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Problem Description
The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+….+ (b)

Input
There are several test cases. Each line has two integers a, b (2< a< b<3000000).

Output
Output the result of (a)+ (a+1)+….+ (b)

Sample Input

3 100

Sample Output

3042

题目大意：求a…b的欧拉函数和
详情参见欧拉函数
注意：要用long long.

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=3000000;
int m[maxn+1],pri[maxn+1],phi[maxn+1];
long long ans;
void euler()
{
    int cnt=0;
    phi[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!m[i])
        {
            pri[++cnt]=m[i]=i;
            phi[i]=i-1;
        }
        for(int j=1;j<=cnt&&pri[j]*i<=maxn;j++)
        {
            m[i*pri[j]]=pri[j];
            if(m[i]==pri[j])
            {
                phi[i*pri[j]]=phi[i]*pri[j];
                break;
            }
            else 
                phi[i*pri[j]]=phi[i]*phi[pri[j]];
        }
    }
}
int main()
{
    euler();
    int a,b;
    while(scanf("%d%d",&a,&b)==2)
    {
        ans=0;
        for(int i=a;i<=b;i++)
            ans+=phi[i];
        printf("%lld\n",ans);
    }
    return 0;
}