hdu 2057 A + B Again

太过爱你忘了你带给我的痛 2022-05-31 09:23 107阅读 0赞

# A + B Again #

**Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 33090 Accepted Submission(s): 13407**

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][Warm Up]

题目解析：

就是让计算十六进制加减法，模拟一下就可以了，非常的简单，注意中间的临时储存要用long long，数比较大，很简单，1A，直接看一下代码把

代码：

#include <bits/stdc++.h>
    using namespace std;
    const int maxn = 10000;
    char c1[maxn],c2[maxn];
    char a[16]= {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    stack <char> sta;
    int main()
    {
        //freopen("in.txt","r",stdin);
        while(cin>>c1>>c2)
        {
            while(!sta.empty())
                sta.pop();
            long long ans =0 ;
            long long flag1=1,flag2=1;
            long long ans1=0,ans2=0;
            long long a1=1,a2=1;
            long long len1 = strlen(c1);
            long long len2 = strlen(c2);
            if(c1[0]=='-')
                flag1 = -1;
            if(c2[0]=='-')
                flag2 = -1;
            for(long long i=len1-1; i>=0; i--)
            {
                if(c1[i]>='0'&&c1[i]<='9')
                {
                    ans1=ans1+a1*(c1[i]-'0');
                    a1*=16;
                }
                if(c1[i]>='A'&&c1[i]<='F')
                {
                    ans1=ans1+a1*(c1[i]-'A'+10);
                    a1*=16;
                }
            }
    
            for(long long i=len2-1; i>=0; i--)
            {
                if(c2[i]>='0'&&c2[i]<='9')
                {
                    ans2=ans2+a2*(c2[i]-'0');
                    a2*=16;
                }
                if(c2[i]>='A'&&c2[i]<='F')
                {
                    ans2=ans2+a2*(c2[i]-'A'+10);
                    a2*=16;
                }
            }
            long long flag = 1;
            ans1*=flag1;
            ans2*=flag2;
            ans=ans1+ans2;
            if(ans<0)
            {
                flag=-1;
                ans=-ans;
            }
            if(ans==0)
            {
                cout<<"0"<<endl;
                continue;
            }
            if(flag==1)
            {
                while(ans)
                {
                    sta.push(a[ans%16]);
                    ans/=16;
                }
            }
            else
            {
                cout<<"-";
                while(ans)
                {
                    sta.push(a[ans%16]);
                    ans/=16;
                }
            }
            while(!sta.empty())
            {
                cout<<sta.top();
                sta.pop();
            }
            cout<<endl;
        }
        return 0;
    }

[Warm Up]: http://acm.hdu.edu.cn/search.php?field=problem&key=%D0%A3%C7%EC%B1%ADWarm+Up&source=1&searchmode=source