HDU 2824 The Euler function --------欧拉模板

The Euler function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2986    Accepted Submission(s): 1221


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
 
第一种打表的方法是,素数和欧拉,分开来打表。250ms
第二种打表只有一个,但是时间上更多。500ms
 
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 
 8 int prime[3000003],len;
 9 int   opl[3000003];
10 bool s[3000003];
11 
12 void Getprime() //打素数表
13 {
14     int i,j;
15     len=0;
16     for(i=2;i<=3000000;i++)
17     {
18         if(s[i]==false)
19         {
20             prime[++len]=i;
21             for(j=i*2;j<=3000000;j=j+i)
22             s[j]=true;
23         }
24     }
25 }
26 
27 void Euler() //欧拉打表。
28 {
29     int i,j;
30     Getprime();
31     for(i=2;i<=3000000;i++)
32     opl[i]=i;
33     opl[1]=0;
34     for(i=1;i<=len;i++)
35     {
36         for(j=prime[i];j<=3000000;j=j+prime[i])
37         opl[j]=opl[j]/prime[i]*(prime[i]-1); //利用的定理
38 
39     }
40 }
41 
42 int main()
43 {
44     int n,m,i;
45     __int64 num;
46     Euler();
47     while(scanf("%d%d",&n,&m)>0)
48     {
49         num=0;
50         for(i=n;i<=m;i++)
51         num=num+opl[i];
52         printf("%I64d\n",num);
53     }
54     return 0;
55 }

 第二种方法。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 
 8 int   opl[3000003];
 9 bool s[3000003];
10 
11 
12 void Euler() //欧拉打表。
13 {
14     int i,j;
15     for(i=2;i<=3000000;i++)
16     opl[i]=i;
17     opl[1]=0;
18 
19     for(i=2;i<=3000000;i++)
20     if(s[i]==false)
21     {
22         for(j=i;j<=3000000;j=j+i)
23         {
24             opl[j]=opl[j]/i*(i-1);
25             s[j]=true;
26         }
27     }
28 }
29 
30 int main()
31 {
32     int n,m,i;
33     __int64 num;
34     Euler();
35     while(scanf("%d%d",&n,&m)>0)
36     {
37         num=0;
38         for(i=n;i<=m;i++)
39         num=num+opl[i];
40         printf("%I64d\n",num);
41     }
42     return 0;
43 }

 

 
posted @ 2013-08-07 17:11  芷水  阅读(156)  评论(0编辑  收藏  举报