The Euler function（线性筛欧拉函数）

时间：2021-01-23 16:49:18
/*

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[]=;

     for(int i=;i<Max;i++)

       euler[i]=i;

     for(int i=;i<Max;i++)

        if(euler[i]==i)

           for(int j=i;j<Max;j+=i)

              euler[j]=euler[j]/i*(i-);//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=;

        for(int i=a;i<=b;i++){

            cur+=euler[i];

        }

        printf("%lld\n",cur);

    }

    return ;

}

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
3 100
 
Sample Output
3042
 
 
Source
2009 Multi-University Training Contest 1 - Host by TJU
 
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