1.求一个n阶方阵对角线元素之和
package week2;import java.util.ArrayList;import java.util.List;import java.util.Scanner;/** * 求一个n阶方阵对角线元素之和 * @author Lenovo_PC */public class Test04 { public static void main(String[] args){ System.out.println("请输入您要计算几阶的方阵(方阵数据由程序给出1到10之间的整数填充):"); Scanner input = new Scanner(System.in); int number = input.nextInt(); List<Integer> list = new ArrayList<Integer>(); for (int i = 1; i <= Math.pow(number, 2); i++) { list.add((int)(Math.random()*11)); } System.out.println("生成的方阵如下所示:"); for (int i = 0; i < list.size(); i++) { System.out.printf("%-2d ",list.get(i)); if ((i+1)%number==0) { System.out.println(); } } int sum = 0, real = number-1; for (int i = real; i < list.size()-real; i+=real) { sum += list.get(i); } System.out.println("主对角线的和为:"+sum); real = number + 1; sum = 0; for (int i = 0; i < list.size(); i+=real) { sum += list.get(i); } System.out.println("副对角线的和为:"+sum); }}
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这题用二维数组更简单直接,第二层for循环判断if(j==i)满足就相加,代码自己写吧,懒得再去敲了
2.输入一个一维数组,最大的与第一个元素交换,最小的与最后一个元素交换,输出数组。
package week2;/** * 输入一个一维数组,最大的与第一个元素交换,最小的与最后一个元素交换,输出数组。 * @author Lenovo_PC */public class Test05 { public static void main(String[] args) { int[] pre = new int[10]; for (int i = 0; i < pre.length; i++) { pre[i] = (int) (Math.random()*100); } System.out.println("自动生成的一个10个元素的数组为:"); for (int i = 0; i < pre.length; i++) { System.out.print(pre[i]+" "); } int max,min,max_temp,min_temp; max=min=pre[0]; max_temp=min_temp=0; for (int i = 1; i < pre.length; i++) { if (max<pre[i]) { max = pre[i]; max_temp = i; } if (min>pre[i]) { min = pre[i]; min_temp = i; } } int t; t = pre[0]; pre[0] = pre[max_temp]; pre[max_temp] = t; t = pre[9]; pre[9] = pre[min_temp]; pre[min_temp] = t; System.out.println(); System.out.println("交换后的数组为:"); for (int i = 0; i < pre.length; i++) { System.out.print(pre[i]+" "); } }}
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这个第二题吧感觉有点抠字眼了,上述代码可以正常的给出结果,但是例如:一个从小打到大排好序的数组[1,2,3,4,5,6,7,8,9,10],这下子问题就来了根据题目的意思最大的与第一个元素交换,最小的与最后一个元素交换,结果是交换两次原样输出,老师的意思是不是输出[10,2,3,4,5,6,7,8,9,1]呢,所以纠结了半天写了下边的程序
package week2;public class Test05_new { public static void main(String[] args) { int[] pre = {1,2,3,4,5,6,7,8,9,10}; for (int i = 0; i < pre.length; i++) { System.out.printf(pre[i]+" "); } int max,min,max_temp,min_temp; max=min=pre[0]; max_temp=min_temp=0; for (int i = 1; i < pre.length; i++) { if (max<pre[i]) { max = pre[i]; max_temp = i; } if (min>pre[i]) { min = pre[i]; min_temp = i; } } int t; if (max_temp == 9 && min_temp ==0) { t = pre[0]; pre[0] = pre[9]; pre[9] = t; }else { t = pre[0]; pre[0] = pre[max_temp]; pre[max_temp] = t; t = pre[9]; pre[9] = pre[min_temp]; pre[min_temp] = t; } System.out.println(); System.out.println("交换后的数组为:"); for (int i = 0; i < pre.length; i++) { System.out.print(pre[i]+" "); } }}
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