Problem 1

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

public int Counnt1(){
		int result = 0;
		int begin =3;
		int end = 1000;
		long t1 = System.currentTimeMillis();
		for(int i=begin; i<end; i++){
			if(i%3==0||i%5==0){
				result += i;
			}
			for_count1++;
		}
		long t2 = System.currentTimeMillis();
		System.out.println(t2-t1);
		return result ;
	}

	public int Counnt2(){
		int result = 0;
		int begin =3;
		int end = 1000;
		long t1 = System.currentTimeMillis();
		for(int i=begin; i<end; i=i+3){
			result += i;
			for_count2++;
		}
		for(int i=5;i<end;i=i+5){
			result += i;
			for_count2++;
		}
		for(int i=15; i<end; i=i+15){
			result -=i;
			for_count2++;
		}
		long t2 = System.currentTimeMillis();
		System.out.println(t2-t1);


		return result;

	}

Output：
0
233168 For_Count:997
0
233168 For_Count:598
方法中的for_count1和for_count2是统计循环次数的！
本来想打印执行时间，可惜。。。
都是0~
不过根据for循环的次数，第二个方法比第一个方法快！