超长位数的字符数的加法与乘法
超长位数的字符数的加法:
测试用例:112233445566778899 + 998877665544332211 = 1111111111111111110
程序代码:
1 #include <stdio.h> 2 #include <string.h> 3 #include <malloc.h> 4 #define MAXLEN 100 5 6 void add(char * a,char * b,char * c) 7 { 8 int i,j; 9 int sa = strlen(a); 10 int sb = strlen(b); 11 int max = sa>sb ? sa : sb; 12 int * s = (int *)malloc(sizeof(int) * (max + 1));//为保证运算和的不溢出,应是最长操作数的位数+1,范围是[0,max]; 13 int * A = (int *)malloc(sizeof(int) * max); 14 int * B = (int *)malloc(sizeof(int) * max); 15 16 for(i=0;i<max;i++) 17 A[i] = B[i] = s[i] = 0; //先初始化为0,防止高位相加时对应位不存在导致的问题 18 s[max] = 0; 19 20 for(i=0;i<sa;i++) //将a倒置以便低位对齐相加 21 A[i] = a[sa - i - 1] - '0'; 22 23 for(i=0;i<sb;i++) 24 B[i] = b[sb - i - 1] - '0'; 25 26 for(i=0;i<max;i++) 27 s[i] = A[i] + B[i]; 28 29 for(i=0;i<max;i++) //集中处理进位问题 30 { 31 if(s[i]>=10) // 若i = max-1时有进位,则s[max] != 0 32 { 33 s[i+1] += s[i] / 10; 34 s[i] %= 10; 35 } 36 } 37 38 if(s[max] != 0) //最高位有进位数据范围为[0,max] 39 { 40 for(j=0;j<=max;j++) 41 { 42 c[j] = s[max - j] + '0'; 43 } 44 c[max+1] = '\0'; 45 } 46 else //最高位无进位,数据范围为[0,max-1] 47 { 48 for(j=0;j<max;j++) 49 { 50 c[j] = s[max -1 - j] + '0'; 51 } 52 c[max] = '\0'; 53 } 54 } 55 56 57 int main() 58 { 59 char a[MAXLEN]; 60 char b[MAXLEN]; 61 char c[2 * MAXLEN]; 62 while(scanf("%s + %s",a,b) != EOF) 63 { 64 add(a,b,c); 65 printf("%s + %s = ",a,b); 66 puts(c); 67 } 68 return 0; 69 }
超长位数的字符数的乘法:
测试用例:112233445566778899 * 998877665544332211 = 112107482103740987777903741240815689
程序代码:
1 #include <stdio.h> 2 #include <string.h> 3 #include <malloc.h> 4 #define MAXLEN 100 5 6 void multiply(char * a,char * b,char * c) 7 { 8 int i,j,ca,cb,* s; 9 ca = strlen(a); //a操作数的位数 10 cb = strlen(b); //b操作数的位数 11 s = (int*)malloc(sizeof(int)*(ca+cb)); //s指向能够存储a和b的空间 12 for (i=0;i<ca+cb;i++) 13 s[i] = 0; //初始化s数组元素全为0 14 for (i=0;i<ca;i++) 15 for (j=0;j<cb;j++) 16 s[i+j+1] += (a[i]-'0') * (b[j]-'0'); 17 18 for (i=ca+cb-1;i>=0;i--) 19 if (s[i]>=10) 20 { 21 s[i-1] += s[i]/10; //高位加上低位的进位 22 s[i] %= 10; 23 } 24 i=0; 25 while (s[i]==0) 26 i++; 27 for (j=0;i<ca+cb;i++,j++) 28 c[j] = s[i] + '0'; 29 c[j]='\0'; 30 free(s); 31 } 32 33 34 int main() 35 { 36 char a[MAXLEN]; 37 char b[MAXLEN]; 38 char c[2 * MAXLEN]; 39 while(scanf("%s * %s",a,b) != EOF) 40 { 41 multiply(a,b,c); 42 printf("%s * %s = ",a,b); 43 puts(c); 44 } 45 return 0; 46 }