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  • 16进制转10进制 HDU-1720

    A+B Coming

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 9461    Accepted Submission(s): 6173


    Problem Description
    Many classmates said to me that A+B is must needs.
    If you can’t AC this problem, you would invite me for night meal. ^_^
     
    Input
    Input may contain multiple test cases. Each case contains A and B in one line.
    A, B are hexadecimal number.
    Input terminates by EOF.
     
    Output
    Output A+B in decimal number in one line.
     
    Sample Input
    1 9 A B a b
     
    Sample Output
    10 21 21
     题意:一个最简单的进制转换问题,只要将两个十六进制的数转化为十进制数相加就好,需注意的是大小写均表示十六进制数。
      思路不用介绍太多,直接上代码说明问题。
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int change(string s1)
     4 {
     5     int num=0,t;
     6     for(int i=0;i<s1.size();i++)
     7     {
     8         if(s1[i]>='0'&&s1[i]<='9')
     9             t=s1[i]-'0';
    10         else if(s1[i]>='A'&&s1[i]<='Z')
    11             t=s1[i]-'A'+10;
    12         else if(s1[i]>='a'&&s1[i]<='z')
    13             t=s1[i]-'a'+10;
    14         num=num*16+t;
    15     }
    16     return num;
    17 }
    18 int main()
    19 {
    20     int sum;
    21     string a,b;
    22     while(cin>>a>>b)
    23     {
    24         sum=change(a)+change(b);
    25         cout<<sum<<endl;
    26     }
    27     return 0;
    28 }

    16进制转10进制,水题。

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  • 原文地址:https://www.cnblogs.com/forever-snow/p/7170917.html
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