大数加法
void jial( )
{
memset( sun,0,sizeof( sun ) );
int len1 = strlen( str1 ),len2 = strlen( str2 );
for( int i = 0; i < ( len1 > len2 ? len1 : len2 ); ++i )//转换成ASCII码
str1[i] -= '0',str2[i] -= '0';
for( int p = 0, q = len1 - 1; p < q; ++p , --q ) //逆序
{
char c = str1[q];
str1[q] = str1[p];
str1[p] = c;
}
for( int p = 0, q = len2 - 1; p < q; ++p,--q)
{
char c = str2[q];
str2[q] = str2[p];
str2[p] = c;
}
for( int i = 0,c = 0; i < ( len1 > len2 ? len1 : len2 ) || c; ++i )//相加 进位
{
if( i < len1 )
c += str1[i];
if( i < len2 )
c += str2[i];
sun[i] = c % 10;//不同的地方1
c /= 10;//不同的地方2
}
}
大数乘法
1 #include "stdio.h"
2 #define MAX 21
3
4 void InputNumber(int []); //输入大数
5 void InitNumber(int []); //初始化用于存放结果的数组
6 void Multi(int [],int [],int []); //进行大数乘法
7 void singleMulti(int [],int,int []); //对大数乘以一位数的函数
8 void Add(int [],int []); //对结果进行错位相加
9 void printresult(int []); //输出结果
10
11 main()
12 {
13 int largenumber1[MAX],largenumber2[MAX],mResult[2 * MAX -1];
14 //largenumber1[]和largenumber2[]表示要相乘的大数
15 //mResult[]表示结果存放的数组，显然其位数不会超过2 * (MAX - 1)
16 printf("Input the first number:\n");
17 InputNumber(largenumber1);
18 printf("Input the second number:\n");
19 InputNumber(largenumber2);
20 InitNumber(mResult);
21 Multi(largenumber1,largenumber2,mResult);
22 printf("The result is:\n");
23 printresult(mResult);
24 getchar();
25 }
26
27 void InputNumber(int largenumber[])
28 {
29 char numberchar;
30 int i = 0;
31 scanf("%c",&numberchar);
32 while(numberchar != '\n' && i < MAX) //最初输入大数用字符串表示
33 {
34 largenumber[++i] = numberchar - '0';
35 scanf("%c",&numberchar);
36 }
37 largenumber[0] = i;
38 }
39
40 void InitNumber(int result[])
41 {
42 for(int i = 0; i < 2 * MAX - 1; i++)
43 result[i] = -1;
44 }
45
46 void Multi(int number1[],int number2[],int result[])
47 {
48 int i;
49 int temper[2 * MAX - 1];
50 for(i = number2[0]; i >= 1; i--)
51 {
52 singleMulti(number1,number2[i],temper);
53 Add(result,temper);
54 }
55 }
56
57 void singleMulti(int number1[],int number2,int temper[])
58 {
59 int i,t = 0,c = 2 * MAX - 2;
60 for(i = number1[0]; i >= 1; i--)
61 {
62 int tempernumber = number2 * number1[i];
63 temper[c--] = (tempernumber + t) % 10;
64 t = (int)((tempernumber + t) / 10);
65 }
66 if (t > 0)
67 {
68 temper[c] = t;
69 temper[0] = 2 * MAX - 2 - c + 1;
70 }
71 else
72 temper[0] = 2 * MAX -2 - c;
73 }
74
75 void Add(int result[],int temper[])
76 {
77 static int pos = 2 * MAX - 2;
78 int c = pos,t = 0;
79 for(int i = 2 * MAX - 2; i > 2 * MAX - 2 - temper[0]; i--)
80 {
81 if (result[c] == -1)
82 result[c] = 0;
83 int tempernumber = result[c] + temper[i];
84 result[c--] = (tempernumber + t) % 10;
85 t = (int)(tempernumber / 10);
86 }
87 if (t == 1)
88 result[c] = 1;
89 pos--;
90 }
91
92 void printresult(int result[])
93 {
94 for(int i = 1; i < 2 * MAX - 1 ; i++)
95 if (result[i] != -1)
96 printf("%d",result[i]);
97 }