A + B Again (16进制的直接运算)
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A + B Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16107 Accepted Submission(s): 6961
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 !
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.
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 121A -9-1A -121A -AA
Sample Output
02C11-2C-90
在十六进制中,除了用0~9外,还用A、B、C、D、E、F分别表示10~15,而十六进制和十进制的不同之处就是在计算时是逢16进一的。如1A(16进制)转化为十进制为1×16+10=26,而214(16进制)转化为十进制为2×16^2+1×16+4=532。(16^2表示16的平方,16^3表示16的3次方,依此类推),而1111(16进制)转化为十进制为1×16^3+1×16^2+1×16+1=4369。 计算的时候也是从低位算起,满16进一,如: 16进制下计算1A+29:最低位A+9即为19,超过16,所以A+9=10+9(十进制)=19(十进制)=13(16进制),高位的1加到前面去,则答案为1A+29=43(16进制)。
HINT:16进制也可以直接加减
#include<stdio.h>int main(){ __int64 a,b; while(scanf("%I64X%I64X",&a,&b)!=EOF) { b=a+b; if(b<0) { b=-b; printf("-"); } printf("%I64X\n",b); } return 0;}
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