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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
    ",ans);
    26     }
    27     return 0;
    28 }
    View Code
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  • 原文地址:https://www.cnblogs.com/lxm940130740/p/3910589.html
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