38 lines
1.2 KiB
Java
38 lines
1.2 KiB
Java
/**
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* 剪绳子，绳子长度乘积最大
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*/
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public class Question14 {
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public static void main(String[] args) {
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int length = 8;
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int result = cutMaxRopeType(length);
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System.out.println(result);
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}
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private static int cutMaxRopeType(int length) {
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//当绳子长度小于4时，因为必须要剪一次（实际不剪断的值已经最大），返回固定计算值。
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if (length < 2) {
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return 0;
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}
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if (length == 2) {
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return 1;
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}
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if (length == 3) {
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return 2;
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}
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/**
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* 当绳子长度大于等于5时，可知3(n-3) > 2(n-2)
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* 所以要把绳子尽量剪成长度为3的段，不够3时剪成长度为2的段
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* 如果把绳子分为大于3的段，比如5=2*3，所以最终还是比较剪成2段和3段的大小
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*/
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int cut3Frequency = length / 3;
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//此时分出一个3，剪成2*2
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if (length - cut3Frequency * 3 == 1){
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cut3Frequency --;
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}
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//剪成2段的次数
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int cut2Frequency = (length - cut3Frequency * 3) / 2;
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int result = (int) (Math.pow(3, cut3Frequency) * Math.pow(2, cut2Frequency));
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return result;
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}
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}
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