一维数组进行多位数算数运算-加减乘

现实中，进行数值很大的数据（超过默认数据类型范围）运算时，可以用数组来完成。

例1：利用一维数组完成三位数加法（N位数加法类似；非最简解法，后同）

 1 int[] a = {0,9,2,3}, b = {0,7,0,9}, c = new int[4];
 2             int jinwei = 0,t;
 3             for (int i = a.Length-1; i >=1; i--)
 4             {
 5                 t = a[i] + b[i];
 6                 c[i] = t % 10+jinwei;
 7                 jinwei = t / 10;
 8             }
 9             c[0] = jinwei;
10             for (int i = 0; i < a.Length; i++)
11             {
12                 Console.Write(a[i]);
13             }
14             Console.Write("+");
15             for (int i = 0; i < b.Length; i++)
16             {
17                 Console.Write(b[i]);
18             }
19             Console.Write("=");
20             for (int i = 0; i < c.Length; i++)
21             {
22                 Console.Write(c[i]);
23             }
24             Console.ReadKey();

运行结果：

三位数减法:

 1 int[] a = { 4, 5, 6 }, b = { 1, 4, 7 }, c= new int[3];
 2             int jw=0,t;
 3             for (int i = a.Length-1; i >=0; i--)
 4             {
 5                 t = a[i] - jw - b[i];
 6                 if (t<0)
 7                 {
 8                     t += 10;
 9                     jw = 1;
10                 }
11                 else
12                 {
13                     jw = 0;
14                 }
15                 c[i] = t;
16             }
17             for (int i = 0; i < c.Length; i++)
18             {
19                 Console.Write(c[i]);
20             }
21             Console.ReadKey();

结果：

例2：三位数乘法

 1 int[] a = { 9, 2, 3 }, b = { 7, 0, 9 }, c = new int[4], d = { 0, 0, 0, 0, 0, 0 };
 2             int jinwei, t;
 3             for (int i = 0; i < b.Length; i++)
 4             {
 5                 jinwei = 0;
 6                 for (int j = 0; j < c.Length; j++)
 7                 {
 8                     c[j] = 0;
 9                 }
10                 for (int j = 0; j < a.Length; j++)
11                 {
12                     t = a[a.Length - 1 - j] * b[b.Length - 1 - i];
13                     c[c.Length - 1 - j] += t % 10;
14                     jinwei = t / 10 + c[c.Length - 1 - j]/10;
15                     c[c.Length - 1 - j] %= 10;
16                     if (jinwei > 0)
17                     {
18                         c[c.Length - 1 - j-1] +=jinwei;
19                     }
20                 }
21                 //测试中间累加的数字
22                 //for (int j = 0; j < c.Length; j++)
23                 //{
24                 //    Console.Write(c[j]);
25                 //}
26                 //Console.WriteLine();
27                 //测试结束
28                 jinwei = 0;
29                 for (int j = 0; j < c.Length; j++)
30                 {
31                     t = d[d.Length - 1 - i - j] + c[c.Length - 1 - j];
32                     d[d.Length - 1 - i - j] = t % 10;
33                     jinwei = t / 10;
34                     jinwei += d[d.Length - 1 - i - j] / 10;
35                     d[d.Length - 1 - i - j] %= 10;
36                     if (jinwei > 0)
37                     {
38                         d[d.Length - 1 - i - j - 1] += jinwei;
39                     }
40                 }
41             }
42             for (int i = 0; i < a.Length; i++)
43             {
44                 Console.Write(a[i]);
45             }
46             Console.Write("*");
47             for (int i = 0; i < b.Length; i++)
48             {
49                 Console.Write(b[i]);
50             }
51             Console.Write("=");
52             for (int i = 0; i < d.Length; i++)
53             {
54                 Console.Write(d[i]);
55             }
56             Console.ReadKey();

运行结果：

利用二位数组完成，会简单一些：

 1 int[] a = { 9, 4, 7 }, b = { 3, 6, 9 };
 2             //M位数和N位数相乘，最多有M+N位数，N位中间结果
 3             int[,] c = new int[b.Length + 1, a.Length + b.Length];
 4             //初始化
 5             for (int i = 0; i < c.GetLength(0); i++)
 6             {
 7                 for (int j = 0; j < c.GetLength(1); j++)
 8                 {
 9                     c[i, j] = 0;
10                 }
11             }
12             //分位乘
13             for (int i = b.Length - 1; i >= 0; i--)
14             {
15                 for (int j = a.Length - 1; j >= 0; j--)
16                 {
17                     c[i, c.GetLength(1) - 1 - (b.Length - 1 - i) - (a.Length - 1 - j)] = a[j] * b[i];
18                 }
19             }
20             //中间结果累加
21             for (int i = 0; i <= c.GetLength(0) - 2; i++)
22             {
23                 for (int j = c.GetLength(1) - 1; j >= 0; j--)
24                 {
25                     c[c.GetLength(0) - 1, j] += c[i, j];
26                 }
27             }
28             //对累加结果的进位部分进行处理
29             for (int i = c.GetLength(1) - 1; i >= 0; i--)
30             {
31                 if (c[c.GetLength(0) - 1, i] >= 10)
32                 {
33                     c[c.GetLength(0) - 1, i - 1] += c[c.GetLength(0) - 1, i] / 10;
34                     c[c.GetLength(0) - 1, i] %= 10;
35                 }
36             }
37             //输出结果
38             for (int i = 0; i < c.GetLength(1); i++)
39             {
40                 Console.Write(c[c.GetLength(0) - 1, i]);
41             }
42             Console.ReadKey();

运行结果：