The Euler function（线性筛欧拉函数）

Posted 2020-08-25 勿忘初心0924

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/*
题意：(n)表示小于n与n互质的数有多少个，给你两个数a，b让你计算a+(a+1)+(a+2)+......+b;

初步思路：暴力搞一下，打表

#放弃：打了十几分钟没打完

#改进：欧拉函数：具体证明看po主的博客 ^0^

#超时：这里直接用欧拉函数暴力搞还是不可以的,用到线性筛欧拉函数，这里总和爆int，要用long long

*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
/**************************欧拉函数模板*****************************/
//筛选法打欧拉函数表 
#define Max 3000010
int euler[Max];
void Init(){ 
     euler[1]=1;
     for(int i=2;i<Max;i++)
       euler[i]=i;
     for(int i=2;i<Max;i++)
        if(euler[i]==i)
           for(int j=i;j<Max;j+=i)
              euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出 
}
/**************************欧拉函数模板*****************************/
int a,b;
int main(){
    // freopen("in.txt","r",stdin);
    Init();
    while(scanf("%d%d",&a,&b)!=EOF){
        ll cur=0;
        for(int i=a;i<=b;i++){
            cur+=euler[i];
        }
        printf("%lld\n",cur);
    }
    return 0;
}

The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 224 Accepted Submission(s): 124

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