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Problem 六

发布时间: 2012-10-25 10:58:57 作者: rapoo

Problem 6
问题描述:
The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.



//求平方和 public long squares_sum(int number){long result = 0;for(int i=1; i<=number; i++){result += i*i;}return result;}//求和的平方public long sum_squares(int number){long result = (number*(number+1))/2;return result*result;}public long count(int number){return sum_squares(number)-squares_sum(number);}

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