目录(?)[+]
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:
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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*/ |
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.
For each line of input produce one line of output. This line contains the value of G for corresponding N.
10 100 500 0
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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: