Huawei Topic Big Data Calculator

2. Large number plus minus

Problem Description:

In the computer, due to the processor bit width limit, can only handle the finite precision decimal integer addition and subtraction, for example, in a 32-bit wide processor computer, the operand and the result of the participating operation must be between -231~231-1. If you need to do a wider range of decimal integer additions, you need to implement it in a special way, such as using a string to save the operand and the result, and to take a bitwise operation. As follows:

9876543210 + 1234567890 =?

Let the string num1= "9876543210", the string num2= "1234567890", and the result is saved in string results = "11111111100".

-9876543210 + (-1234567890) =?

Let the string num1= "-9876543210", the string num2= "1234567890", and the result is saved in the string results = "11111111100".

Requires programming to achieve the above-mentioned high-precision decimal addition.

Required implementation functions:

void Add (const char *NUM1, const char *num2, char *result)

"Input" NUM1: string as operand 1, if the operand is negative, then num1[0] is the sign bit '-'; num2: String form operand 2, if the operand is negative, then num2[0] is the sign bit '-'

Output Result: Saves the addition calculation result string, or result[0] as the sign bit if the result is negative.

Note:

I, when the input is positive, ' + ' will not appear in the input string, when the input is negative, '-' will appear in the input string, and must be in the leftmost position of the input string;

II, input string all bits represent a valid number, that is, there is no input string starting from ' 0 ', such as "0012", "0012" will not appear;

III, requires that the output string all bits are valid numbers, the result is positive or 0 o'clock ' + ' does not appear in the output string, the result is negative when the leftmost position of the output string is '-'.

Example

Input: NUM1 = "580"

num2 = "-50"

Output: result = "530"

Input: NUM1 = "580"

num2 = "-600"

Output: result = "-20"

#include <stdio.h> #include <string.h>//"input" NUM1: the string form operand 1, if the operand is negative, then num1[0] is the sign bit '-'; num2: string as operand 2, If the operand is negative, then num2[0] is the sign bit '-' output ' result: Saves the addition calculation results string, and if the result is negative, then result[0] is the sign bit. void Change (char * str) {int num =strlen (str); int i; int tmp;for (i = 0;i <= (NUM/2-1); i++) {tmp = Str[i];str[i] = str[ Num-i -1];str[num-1-i] = tmp;} }void Jinwei (char *str) {int num = strlen (str), if (str[num-1]! = '-') {for (int i = 0;i<num;i++) {if (Str[i] >= 0x3a) {str[ i+1]++;str[i]= Str[i]-10;if (i = = num-1) {str[i+1] = ' 1 ';}}} else{for (int i = 0;i<num-1;i++) {if (Str[i] >= 0x3a) {str[i+1]++;str[i]= str[i]-10;if (i = = num-2) {str[i+1] = ' 1 '; str [I+2] = '-';}}}} void Jinwei2 (char *str) {int num = strlen (str); for (int i =0;i<num-1;i++) {if (str[i]<0x30) {str[i+1]--;str[i]=str[i ]+10;}}} void Add (const char *num1,const char *num2, char *result) {char num11[50] = ""; char num22[50] = ""; strcpy (NUM11,NUM1); STRCP Y (num22,num2), change (NUM11), change (NUM22), int i1 = strlen (NUM11); int i2 = strlen (NUM22); int imin; int i;int flag;int t;if (num1[0]! = '-' && num2[0]! = '-') {I1 >= i2? (Imin = i2): (imin = i1);//Get a small string of I1 >= i2 in the plus number?  (flag = 2): (flag = 1); Flag bit which is the smaller string for (i=0;i<imin;i++) {result[i] = Num11[i] + num22[i]-0x30;} if (flag = = 2) for (; i<i1;i++) Result[i] = Num11[i];else if (flag = = 1) for (; i<i2;i++) result[i] = Num22[i];} if (num1[0] = = '-' && num2[0] = = '-') {I1 >= i2? (Imin = i2): (imin = i1);//Get a small string of I1 >= i2 in the plus number?  (flag = 2): (flag = 1); Flag bit which is the smaller string for (i=0;i<imin-1;i++) {result[i] = Num11[i] + num22[i]-0x30;} if (flag = = 2) for (; i<i1;i++) Result[i] = Num11[i];else if (flag = = 1) for (; i<i2;i++) result[i] = Num22[i];} if (num1[0]! = '-' && num2[0] = = '-') {change (NUM11); change (NUM22); flag = strcmp (NUM11, (num22+1)); if (Flag > 0) /positive number greater than negative {change (NUM11), change (NUM22), for (i =0;i< i1;i++) {if (i<i2-1) result[i] = Num11[i]-num22[i] + ' 0 '; Elseresult[i] = Num11[i];} if (flag < 0)//positive is less than negative {change (NUM11), change (NUM22), for (i= 0;i<i2;i++) {if (I&LT;I1) result[i] = Num22[i]-num11[i] + ' 0 '; elseresult[i] = Num22[i];}} if (flag = = 0) result[0] = ' 0 ';} if (num1[0] = = '-' && num2[0]! = '-') {change (NUM11); change (NUM22); flag = strcmp ((num11+1), NUM22); if (Flag > 0) /positive Xiao to negative {change (NUM11), change (NUM22), for (i = 0;i<i1;i++) {if (I&LT;I2) result[i] = Num11[i]-num22[i] + ' 0 '; Elseresult[i] = Num11[i]; }} if (flag < 0)//positive number da is negative {change (NUM11), change (NUM22), for (i =0;i< i2;i++) {if (i<i1-1) result[i] = Num22[i]- Num11[i] + ' 0 '; elseresult[i] = Num22[i];} } if (flag = = 0) result[0] = ' 0 ';}} void Qianyi (char *str) {int I =0;for (i = 1;i<strlen (str); i++) {str[i-1] = Str[i];}} void Xianshi (char *str) {int i;if (str[0] = = '-') for (i = 1;i<strlen (str);) {if (str[i] = = ' 0 ') {Qianyi ((str+i)); str[ strlen (str)-1] = ' + ';} Elsebreak;} Elsefor (i = 0;i<strlen (str); i++) {if (str[i] = = ' 0 ') Qianyi ((str+i)); elsebreak;}} int main () {char a1[] = "580"; char a2[] = " -600"; char a3[50] = "";//char a3[]; Add (A1,A2,A3); Change (A1);if (A1[0]!=a2[0]) jinwei2 (A3); Elsejinwei (A3); Change (A3); Xianshi (A3);p rintf ("%s", A3);}

　　

Huawei Topic Big Data Calculator