hdoj 2824 The Euler function 【欧拉函数】

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4520    Accepted Submission(s): 1880

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

 

分析：

题意就是求出从n~m的数的质数的个数的和。

代码：

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define maxn 3000001using namespace std;int n,m;int gg[maxn];void euler(){     memset(gg,0,sizeof(gg));     gg[1]=1;     for(int i=2;i<maxn;i++)     if(!gg[i])     for(int j=i;j<maxn;j+=i)      {       if(!gg[j])       gg[j]=j;       gg[j]=gg[j]/i*(i-1);      }}int main(){      euler();// 这一点特别注意，不要把函数调用放在while里面 否则会超时。     while(scanf("%d%d",&n,&m)!=EOF)    {     long long sum=0;// 此处用long long 和__int64 都可以。      for(int i=n;i<=m;i++)     {      if(gg[i])      sum+=gg[i];     }     printf("%lld\n",sum);    }    return 0;}