1. 设计一个类（不妨叫做A ），实现对一维数组的指定范围内的元素相加运算。
import java.util.Scanner;
public class A {
public static void main(String[] args) {
int[] aa = { -2574, -4010, 8243, 771, 2447, -5197, 2556, 8044, 3314,
3617, 6065, -2817, 3131, 6318, 2186, -113, 629, -2582, -37,
-1520, 164, 2055, -5936, 5912, 1717, 5988, 4781, 5757, 892,
-4394, 8034, 2213, -1080, -2080, 5364, 106, 2657, 566, 3940,
-5116, 4583, 1806, 6555, 2621, -7197, 528, 1626, 18, 1049,
6243, 3198, 4397, -1325, 9087, 936, -6291, 662, -178, 135,
-3473, -2385, -165, 1713, -7949, -4234, 1138, 2212, 104, 6968,
-3632, 3801, 1137, -1296, -1215, 4272, 6223, -5922, -7723,
7044, -2938, -8180, 1356, 1159, -4022, -3713, 1158, -8715,
-4081, -2541, -2555, -2284, 461, 940, 6604, -3631, 3802, -2037,
-4354, -1213, 767 };//
int sum = 0;
Scanner a = new Scanner(System.in);
System.out.println("从第几个开始(<=100)：");
int start = a.nextInt();
System.out.println("到第几个结束(<=100)：");
int end = a.nextInt();
System.out.println("第" + start + "个元素到第" + end + "个元素的和为:");
for (int i = start - 1; i < end; i++)
// 第一个是aa[0]
sum += aa[i];
System.out.println(aa[start - 1] + "+...+" + aa[end - 1] + "=" + sum);
}
}
运行结果为：开始数1~结束数10的元素相加为17211
2. 设计一个类（不妨叫做B ），实现对二维数组(a[N][M])的指定范围内的元素相加运算。
import java.util.Scanner;
public class B {
public static void main(String[] args) {
int[][] b = {
{ -2574, -4010, 8243, 771, 2447, -5197, 2556, 8044, 3314, 3617,
6065, -2817, 3131, 6318, 2186, -113, 629, -2582, -37,
-1520, 164, 2055, -5936, 5912, 1717, 5988, 4781, 5757,
892, -4394, 8034, 2213, -1080, -2080, 5364, 106, 2657,
566, 3940, -5116, 4583, 1806, 6555, 2621, -7197, 528,
1626, 18, 1049, 6243, 3198, 4397, -1325, 9087, 936,
-6291, 662, -178, 135, -3473, -2385, -165, 1713, -7949,
-4234, 1138, 2212, 104, 6968, -3632, 3801, 1137, -1296,
-1215, 4272, 6223, -5922, -7723, 7044, -2938, -8180,
1356, 1159, -4022, -3713, 1158, -8715, -4081, -2541,
-2555, -2284, 461, 940, 6604, -3631, 3802, -2037,
-4354, -1213, 767 },
{ 2740, -4182, -5632, -2966, -1953, 567, -8570, 1046, 2211,
1572, -2503, -1899, 3183, -6187, 3330, 3492, -464,
-2104, 316, 8136, 470, 50, 466, -1424, 5809, 2131,
6418, -3018, 6002, -8379, 1433, 1144, 2124, 1624, -602,
-5518, 5872, 870, -5175, -3961, -427, -6284, 2850, 481,
6175, 141, -766, -1897, -748, -4248, 366, 4823, 3003,
1778, 3256, 2182, 2253, 5076, 5540, -2650, 2451, -1875,
5482, -6881, -329, -969, -8032, -2093, 612, 1524,
-5492, 5758, -7401, -5039, 3241, 6338, 3581, 4321,
-1072, 4942, 2131, 210, -7045, -7514, 7450, -1142,
-2666, -4485, -639, 2121, -5298, -3805, -1686, -2520,
-1680, 2321, -4617, -1961, 2076, 7309 } };
int sum=0;
Scanner bb = new Scanner(System.in);
System.out.println("开始列:");
int startI = bb.nextInt();
System.out.println("结束列:");
int endI = bb.nextInt();
System.out.println("开始行:");
int startJ = bb.nextInt();
System.out.println("结束行:");
int endJ = bb.nextInt();
System.out.println("开始列" + startI + "结束列" + startJ + "开始行" + endI + "结束行"
+ endJ + "和为：");
for (int i = startI - 1; i < endI; i++)
{
for (int j = startJ - 1; j < endJ ; j++)
{
sum += b[i][j];
}
}
System.out.println(b[startI - 1][startJ - 1] + "+...+"
+ b[endI - 1][endJ - 1] + "=" + sum);
}
}
运行结果：
开始列：
1
结束列：
2
开始行：
1
结束行：
39
开始列1结束列2开始行1结束行39和为：
60441
这次的作业看似很复杂，其实做起来更麻烦，起初毫无头绪，但是经过多天和同学们交流之后终于有了起色，花了多天的研究跟思考，终于打破了我内心的恐惧，写了一点点东西，虽然我知道并不好，但我会继续努力。（ps：由于电脑问题，插图比较麻烦，所以就纯手打了，请老师务必见谅）