HDU-1002 A + B Problem II

Posted 2020-11-25 bethebestone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 515227    Accepted Submission(s): 98635

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

 

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

 

解题思路：

如果同学们会Java的话，就可以直接用大数进行计算

 

但是练习一下C++的大数加法还是挺好的

这道题C++不能直接进行大数的加法，所以我采用string数组来进行辅助计算，会比较简单一些

利用string的相关函数进行操作会比较便捷

还有就是一些加法上的小细节了，注意进位之类的，从后往前加，然后还要注意有没有加完

 

代码如下：

 1 #include<bits/stdc++.h>
 2 #define mem(a) memset(a,0,sizeof(a))
 3 #define forn(i,n) for(int i=0;i<n;++i)
 4 #define for1(i,n) for(int i=1;i<=n;++i)
 5 #define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
 6 using namespace std;
 7 typedef long long ll;
 8 const int maxn=1e3+5;
 9 const int mod=10007;
10 const int inf=0x3f3f3f3f;
11 
12 ll n,m,t,k;
13 string num1,num2;
14 
15 string add(string stra,string strb)
16 {
17     string res;
18     string str1=stra;
19     string str2=strb;
20     reverse(str1.begin(),str1.end());
21     reverse(str2.begin(),str2.end());
22     int i=0;
23     int x=0;
24     while(i<str1.size()&&i<str2.size())
25     {
26         int a=str1[i]-‘0‘;
27         int b=str2[i]-‘0‘;
28         int c=a+b+x;
29         x=c/10;
30         c=c%10;
31         char s=‘0‘+c;
32         res=s+res;
33         i++;
34     }
35     if(i<str1.size())
36     {
37         while(i<str1.size())
38         {
39             str1[i]=str1[i]+x;
40             int w=str1[i]-‘0‘;
41             x=w/10;
42             w=w%10;
43             str1[i]=‘0‘+w;
44             res=str1[i]+res;
45             i++;
46         }
47     }
48     if(i<str2.size())
49     {
50         while(i<str2.size())
51         {
52             str2[i]=str2[i]+x;
53             int w=str2[i]-‘0‘;
54             x=w/10;
55             w=w%10;
56             str2[i]=‘0‘+w;
57             res=str2[i]+res;
58             i++;
59         }
60     }
61     if(x==1)
62     {
63         char cc=‘1‘;
64         res=cc+res;
65     }
66     return res;
67 }
68 
69 int main()
70 {
71     IO;
72     cin>>m;
73     int k=1;
74     while(k<=m)
75     {
76         if(k!=1)
77             cout<<endl;
78         cin>>num1>>num2;
79         string ans=add(num1,num2);
80         cout<<"Case "<<k<<":"<<endl;
81         cout<<num1<<" + "<<num2<<" = "<<ans<<endl;
82         k++;
83     }
84     return 0;
85 }