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hdu2824（欧拉函数）
The Euler function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3674 Accepted Submission(s): 1509
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
总结：最初，我用sum[i]数组存储1到i的欧拉函数值得和，但是因为开了两个数组，超了空间，本想用此法优化时间的，结果还是要从a循环到b取每个数的欧拉函数的值相加
1 #include<stdio.h> 2 #include<string.h> 3 __int64 euler[3000000]; 4 int main() 5 { 6 __int64 ans; 7 memset(euler,0,sizeof(euler)); 8 euler[1] = 1; 9 int a,b,i,j;10 for(i = 2; i <3000000; i++)11 {12 if(!euler[i])13 for(j = i; j <3000000; j += i)14 {15 if(!euler[j])16 euler[j] = j;17 euler[j] = euler[j]/i*(i-1);18 }19 }20 while(scanf("%d%d",&a,&b)!=EOF)21 {22 ans=0;23 for(i=a; i<=b; i++)24 ans+=euler[i];25 printf("%I64d\n",ans);26 }27 return 0;28 }
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