A + B Again

最新推荐文章于 2020-06-20 16:13:13 发布

原创最新推荐文章于 2020-06-20 16:13:13 发布 · 358 阅读

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A + B Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28319 Accepted Submission(s): 11488

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

Author

linle

Source

校庆杯Warm Up

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#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include <set>
#include<algorithm>
#include <limits.h>
#include <ctype.h>
#include <map>
#include <stack>
#include <sstream>
typedef long long LL;
using namespace std;
const int N = 105;
int data[N][N];
int dp[N][N];
char c[100][100] = {' '};
#define PI 3.1415926
typedef __int64 LL;
int main() {
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //涉及到16进制内的加法，
    //可以用%I64x直接来处理，
    //要注意到16进制中负数是用补码来表示的。
    //一个比较困惑的事实是，
    //这道题再输出时，
    //%64X中‘X’必须是大写，小写是过不了的。
    LL a, b;
    while(scanf("%I64X %I64X", &a, &b) != EOF){
        LL ans = a + b;
        if(ans < 0){ //16进制没有负数输出
            printf("-");
            ans = -ans;
        }
        printf("%I64X\n", ans);
    }
    return 0;
}