HDUACM-2057 十六进制相加

A + B Again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21410    Accepted Submission(s): 9250

Problem Description

There must be many A + B problems in our HDOJ , now a new one is coming.
Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too.
Easy ? AC it !

 

Input

The input contains several test cases, please process to the end of the file.
Each case consists of two hexadecimal integers A and B in a line seperated by a blank.
The length of A and B is less than 15.

 

Output

For each test case,print the sum of A and B in hexadecimal in one line.

 

Sample Input

  

   +A -A
+1A 12
1A -9
-1A -12
1A -AA
  

 

Sample Output

  

   0
2C
11
-2C
-90
  

 

Author

linle

 

Source

校庆杯Warm Up

 

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#include<stdio.h>
#include<String.h>
int toHex(__int64	a,char s[]){
	int index = 0,l;
	memset(s,0,sizeof(s));
	
	if(a<0){
		a = -a;
		index = 1;
		s[0] = '-';
	}
	
	do{
		l = a%16;
		if(l>9){
			s[index++] = 'A' + l -10;
		}else{
			s[index++] = '0' + l ;
		}
		
		a = a/16;
	}while(a!=0);
	return index;         //此处如不返回index表示数组元素个数的话  数组长度会不对  我也不知道为什么....

}


__int64 toDec(char s[]){
	char sy[2] = {0};   //符号位 
	int index = 0,len = strlen(s);
	__int64 c = 1,sum = 0;    //C不断乘16  sum结果 
	if(s[0] == '-'||s[0] == '+'){
		sy[0] = s[0];index = 1;
	}
	for(int i = len-1;i >=index;i--){
		if(s[i]>='A'&&s[i]<='F'){
			sum += ((int)(s[i] - 55))*c;    //大写英文字符 
		}else{
			sum += (s[i]-48)*c;     //数字字符 
		
		} 
		c = c*16; 
		
	}
	//printf("%I64d\n",c);
	//printf("%I64d\n",sum);
	if(sy[0]=='-'){
		return -sum;
	}else{
		return sum;
	} 
}
int main(){
	
	char a[20],b[20],c[20];
	__int64 add,len;
	while(scanf("%s%s",a,b)!=EOF){
	
		add = toDec(a)+toDec(b);
		len = toHex(add,c);
	
		if(c[0]=='-'){
			printf("-");
			for(int i = len -1 ;i >= 1 ;i--){
				printf("%c",c[i]);
			}
		}else{
			for(int i = len -1 ;i >= 0 ;i--){
				printf("%c",c[i]);
			}
		}
		
		printf("\n");
	} 
	
	return 0;
}