Project Euler - 6

欧拉项目 第6题:

problem :

The sum of the squares of the first ten natural numbers is,

1 2 + 2 2 + ... + 10 2 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10) 2 = 55 2 = 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: 1到10所有数的平方和等于385, 1到10所有数的和的平方等于3025,两个数的差为 2640。 求1到100的结果。


package projectEuler;

public class Problem6 {
	public static void main(String[] args){
		long sum1 = 0;
		long sum2 = 0;
		for(int i=1; i<=100; i++){
			sum1 += i*i;
		}
		
		int temp = 0;
		for(int i=1; i<=100; i++){
			temp += i;
		}
		sum2 = temp*temp;
		
		System.out.println(""+(sum2-sum1));
	}
}


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