problem 23

问题描述：

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

<!-- <p>A number whose proper divisors are less than the number is called deficient and a number whose proper divisors exceed the number is called abundant.</p> -->

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

解决方法：

 

	public static int sum_divisors(int number){
	
		int result = 1;
		if(number%2!=0){
			for(int i=3; i*i<=number; i+=2){
				if(number%i==0){
					result +=i;
					int other_i = number/i;
					if(i!=other_i&&number%other_i==0){
						result +=other_i;
					}
				}
			}
		}else{
			int two = 2;
			int step =0;
			int divide =1;
			int original = number;
			while(number%2==0){
				number = number/2;
				divide *=2;
				step++;
			}
			if(number==1){
				step--;
			}
			for(int i=0; i<step; i++){
				result += two;
				two *=2;
			}
			number = original/divide;
//			System.out.println(step+"_"+number+"_"+result);
			for(int i=3; i<=number; i+=2){
				if(number%i==0){
					result +=i;
					int begin = 2;
					for(int j=0; j<step; j++){
						if(i*begin<original){
							result += i*begin;
							begin *=2;
						}else{
							break;
						}
					}
				}
			}
		}
		
		return result ;
	}
	
	public static int total(){
		int result = 0;
		
		Set<Integer> abundant = new HashSet<Integer>();
		
		for(int i=4; i<UPPER; i++){
			if(sum_divisors(i)>i)
				abundant.add(i);
		}
		
		for(int i=1; i<UPPER; i++){
			int middle = i/2;
			boolean ok = true;
			for(int j=4; j<=middle; j++){
				if(abundant.contains(j)&&abundant.contains(i-j)){
					ok = false;
					break;
				}
			}
			if(ok){
				result += i;
			}
		}
		return result;
	}

 

评论

1 楼 to_zoe_yang 2011-08-07  

这道题还能继续优化，计算除数之和还能改进！