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Java进阶之欧拉工程 第六篇【持续更新】

2014-08-15 12:55 316 查看
原题如下:

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.
翻译:
前10个自然数的平方和为385,前十个自然数的和平方为3025,它们的差值为3025-385=2640,要求前一百个自然数的和平方与平方和的差值,这道题就很简单了,代码如下

public class Launcher {

public static void main(String[] args) {

int num1=0,num2=0;
for(int i =1;i<101;i++){
num1=(int) (num1+Math.pow(i,2));
num2+=i;
}

int answer=(int) (Math.pow(num2,2)-num1);
System.out.println(answer);

}

}
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