hdu2824 The Euler function（欧拉函数个数）

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vasttian https://blog.csdn.net/u012860063/article/details/36426357

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2824

欧拉函数性质：

1：（百科）：http://baike.baidu.com/link?url=r-yneKCCyS9N6bhbQCqiZX0V2OCYq9r7iHSzHTSs03H7qRvu1OfUzlOxfVEs2PmR

2：http://www.cppblog.com/doer-xee/archive/2009/12/01/102353.html

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

题意：求a到b中每一个数的欧拉函数的总和！

代码例如以下：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#define N 3000001

__int64 phi[N];

void init()
{
    int i, j;
    for(i = 1; i < N; i++)
        phi[i] = i;

    for(i = 2; i < N; i++)
        if(i == phi[i]) //此时i为素数
            for(j = i; j < N; j += i)  //j累加i
                phi[j] = (phi[j] / i) * (i - 1); //j有因子i,并且i是素数,正是欧拉函数
}

int main()
{
    init();
    int a, b;
    while(scanf("%d%d", &a, &b) != EOF)
    {
        __int64 ans = 0;
        for(int i = a; i <= b; i++)
            ans += phi[i];
        printf("%I64d\n", ans);
    }
    return 0;
}