[英]Algorithm to find `balanced number` - the same number of even and odd dividers
我们将平衡数定义为具有相同数量的偶数和奇数除法器的数,例如(2 和 6 是平衡数)。 我试图为波兰 SPOJ 做任务,但我总是超过时间。 任务是找到大于输入给定的最小余额数。 有示例输入:
2 (amount of data set)
1
2
和 output 应该是:
2
6
这是我的代码:
import java.math.BigDecimal;
import java.util.Scanner;
public class Main {
private static final BigDecimal TWO = new BigDecimal("2");
public static void main(String[] args) throws java.lang.Exception {
Scanner in = new Scanner(System.in);
int numberOfAttempts = in.nextInt();
for (int i = 0; i < numberOfAttempts; i++) {
BigDecimal fromNumber = in.nextBigDecimal();
findBalancedNumber(fromNumber);
}
}
private static boolean isEven(BigDecimal number){
if(number.remainder(new BigDecimal("2")).compareTo(BigDecimal.ZERO) != 0){
return false;
}
return true;
}
private static void findBalancedNumber(BigDecimal fromNumber) {
BigDecimal potentialBalancedNumber = fromNumber.add(BigDecimal.ONE);
while (true) {
int evenDivider = 0;
int oddDivider = 1; //to not start from 1 as divisor, it's always odd and divide potentialBalancedNumber so can start checking divisors from 2
if (isEven(potentialBalancedNumber)) {
evenDivider = 1;
} else {
oddDivider++;
}
for (BigDecimal divider = TWO; (divider.compareTo(potentialBalancedNumber.divide(TWO)) == -1 || divider.compareTo(potentialBalancedNumber.divide(TWO)) == 0); divider = divider.add(BigDecimal.ONE)) {
boolean isDivisor = potentialBalancedNumber.remainder(divider).compareTo(BigDecimal.ZERO) == 0;
if(isDivisor){
boolean isEven = divider.remainder(new BigDecimal("2")).compareTo(BigDecimal.ZERO) == 0;
boolean isOdd = divider.remainder(new BigDecimal("2")).compareTo(BigDecimal.ZERO) != 0;
if (isDivisor && isEven) {
evenDivider++;
} else if (isDivisor && isOdd) {
oddDivider++;
}
}
}
if (oddDivider == evenDivider) { //found balanced number
System.out.println(potentialBalancedNumber);
break;
}
potentialBalancedNumber = potentialBalancedNumber.add(BigDecimal.ONE);
}
}
}
它似乎工作正常,但太慢了。 你能帮忙找到优化它的方法吗,我错过了什么吗?
正如@MarkDickinson 建议的那样,答案是:
private static void findBalancedNumberOptimized(BigDecimal fromNumber) { //2,6,10,14,18,22,26...
if(fromNumber.compareTo(BigDecimal.ONE) == 0){
System.out.println(2);
}
else {
BigDecimal result = fromNumber.divide(new BigDecimal("4")).setScale(0, RoundingMode.HALF_UP).add(BigDecimal.ONE);
result = (TWO.multiply(result).subtract(BigDecimal.ONE)).multiply(TWO); //2(2n-1)
System.out.println(result);
}
}
最后是绿色的,谢谢马克!
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