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1002 大数相加

2024-07-05 15:22:49   229人阅读

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
 
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
 
Sample Input
2 1 2 112233445566778899 998877665544332211
 
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
 
难点是，数据类型最长有32位（4字节或者2字），数值范围是-2147483648~2147483648或者0~4294967295，但题目中指出输入数据位数长度可以达到1000位，10^999>>4294967295,故不能用常规方法
 
具体解决方法是，将数字利用字符串的形式表示，每个字符都是数字，1000个连续字符也没问题，再将两个不同字符串相加得到最终结果。
 
有一次提交时，出现了“Presentation Error”的错误，缘由是输出结果的格式不符合要求，比方少个空格什么的。
 1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 int main()
 5 {
 6     int n;
 7     while(cin>>n)//n为case数 
 8  {
 9     for(int i=1;i<=n;i++)
10     {
11         string a,b,c;//3个字符串 
12         cin>>a>>b;
13         int la=a.length()-1,lb=b.length()-1,jw=0,ta,tb,tt,f=0;
14         char tc;
15         while(la>=0||lb>=0)
16         {
17         
18             if(la<0) ta=0;
19                 else ta=a[la]-‘0‘;
20             if(lb<0) tb=0;
21                 else tb=b[lb]-‘0‘;
22             tt=jw+ta+tb;//tt为a和b两位相加结果 
23             jw=tt/10;//jw为进位
24             tc=tt%10+‘0‘;//tc为赋值给字符串c之前的一个中转
25             if(tc!=‘0‘) f=1;//f为进位标志
26             c+=tc;
27             la--;lb--;
28         }
29         if(jw>0)
30         {
31             f=1;
32             tc=jw+‘0‘;
33             c+=tc;
34         }
35 
36         if(i!=1) cout<<endl;
37         cout<<"Case "<<i<<":"<<endl;
38         cout<<a<<" + "<<b<<" = ";
39         if(f==1)
40         {
41             for(int j=c.length()-1;j>=0;j--)
42                 cout<<c[j];
43             cout<<endl;
44         }
45         else  cout<<0<<endl;
46     }
47     }
48     return 0;
49 }
 


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