UVa 11417 GCD (欧拉φ函数)

 

UVa 11417 GCD (欧拉φ函数)

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目录(?)[+]

11417 - GCD

Time limit: 2.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2412

Given the value of N, you will have to find the value of G. The definition of G is given below:

 

Here GCD(i,j) means the greatest common divisor of integer i and integer j.

 

For those who have trouble understanding summation notation, the meaning of G is given in the following code:

G=0;

for(i=1;i<N;i++)

for(j=i+1;j<=N;j++)

{

    G+=GCD(i,j);

}

/*Here GCD() is a function that finds the greatest common divisor of the two input numbers*/

 

Input

The input file contains at most 100 lines of inputs. Each line contains an integer N (1<N<501). The meaning of N is given in the problem statement. Input is terminated by a line containing a single zero.  This zero should not be processed.

 

Output

For each line of input produce one line of output. This line contains the value of G for corresponding N.

 

Sample Input                 Output for Sample Input

10

100

500 

0                                      

 

67

13015

442011                                   


如何求?

思路:可以直接算,复杂度O(N^2 logN),但是我们可以找到一种复杂度更小的算法O(N loglogN)

以10为例,与之互素的有φ(10)=4个(1,3,7,9),与之gcd=2的有φ(10/2)=4个(2,4,6,8),与之gcd=5的有φ(10/5)=1个(5)

这样,10提供的G值就是5*φ(2)+2*φ(5)+φ(10)=5+8+4=17

根据上述计算过程可以得到如下公式:


(方括号指艾弗森约定,当方括号内语句为真时其值为1,假时为0,参见《具体数学》P21)


完整代码:

[cpp]  view plain copy
  1. /*0.013s*/  
  2.   
  3. #include<cstdio>  
  4. const int maxn = 501;  
  5.   
  6. int phi[maxn], G[maxn];  
  7.   
  8. void init()  
  9. {  
  10.     int i, j;  
  11.     for (i = 2; i < maxn; ++i)  
  12.         phi[i] = i;  
  13.     for (i = 2; i < maxn; ++i)  
  14.     {  
  15.         if (phi[i] == i)  
  16.             for (j = i; j < maxn; j += i)  
  17.                 phi[j] = phi[j] / i * (i - 1);///计算欧拉φ函数  
  18.         for (j = 1; j * i < maxn; ++j)  
  19.             G[j * i] += j * phi[i];  
  20.     }  
  21.     for (i = 3; i < maxn; ++i)  
  22.         G[i] += G[i - 1];  
  23. }  
  24.   
  25. int main()  
  26. {  
  27.     init();  
  28.     int n;  
  29.     while (scanf("%d", &n), n)  
  30.         printf("%d\n", G[n]);  
  31.     return 0;  
  32. }  

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