HDU2057-A + B Again
阿新 • • 發佈:2018-11-03
HDU2057-A + B Again
題目:
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
#include<bits/stdc++.h> using namespace std; int main() { long long a,b; while(~scanf("%llX%llX",&a,&b)) { if(a + b < 0) { a = -a; b = -b; cout << "-"; } printf("%llX\n",a + b); } return 0; }
需要注意的是如果用long long,輸入和輸入時都需要寫成 %llX ,%x和%X分別對應十六進位制中字母小寫和字母大寫
另外如果相加出現負數的時候需要在前面手動加“-”,然後把a,b分別取反,如果不加就會跳出一堆數字。。。。