zoukankan      html  css  js  c++  java
  • hdu 2824 The Euler function

    The Euler function

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 5028    Accepted Submission(s): 2123


    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

     题意:求a到b之间的每个数的欧拉函数的和

    #include<stdio.h>
    #include<string.h>
    #include<cstdio> 
    #include<string>
    #include<math.h>
    #include<algorithm>
    #define LL long long
    #define PI atan(1.0)*4
    #define DD double
    #define MAX 3000010
    #define mod 100
    #define dian 1.000000011
    #define INF 0x3f3f3f
    using namespace std;
    LL s[MAX],a[MAX];
    void el()
    {
    	LL i,j;
    	for(i=2;i<MAX;i++)
    	{
    		if(!a[i])
    		{
    			for(j=i;j<MAX;j+=i)
    			{
    				if(!a[j]) a[j]=j;
    				a[j]=a[j]/i*(i-1);
    			}
    		}
    		a[i]+=a[i-1];
    	}
    }
    int main()
    {
    	LL n,m,j,i,t;
    	el();
    	while(scanf("%lld%lld",&n,&m)!=EOF)
    	{
    		t=a[m]-a[n-1];
    		printf("%lld
    ",t);
    	}
    	return 0;
    } 
    

      

  • 相关阅读:
    第五次作业之例行报告
    欢迎来怼—第三次Scrum会议
    例行报告
    探路者 Alpha阶段中间产物
    第六次作业—例行报告
    第五次作业—例行报告
    第四次作业—单元测试
    第四次作业—例行报告
    第四次作业—结对编程 四则运算
    第四次作业—1.代码规范
  • 原文地址:https://www.cnblogs.com/tonghao/p/5316932.html
Copyright © 2011-2022 走看看