A + B Problem II（大数加法）

http://acm.hdu.edu.cn/showproblem.php?pid=1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 261608    Accepted Submission(s): 50625

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

题解，大数加法一般可以用java，或者用string 或者用数组手动模拟算法，用c++的时候最好用模板

现在先给出java大数加法的代码

 import java.math.BigInteger;
 import java.util.Scanner;

 public class Main {
     public static void main(String[] args) {
         Scanner cin = new Scanner(System.in);//大数的输入，定义一个输入器
         BigInteger a = null, b = null, c = null;//开始要赋值成空
         a = BigInteger.valueOf();
         b = BigInteger.valueOf();
         int T;
         T = cin.nextInt();//读入T;
 //        while(cin.hasNextBigInteger())//判断是否读到文件结尾相当于while(~scanf())
         for(int cas = ; cas <= T; cas++)
         {
             a = cin.nextBigInteger();
             b = cin.nextBigInteger();

 //            BigInteger zero = BigInteger.valueOf(0);//大数判断是不是等于0
 //            if(a.equals(BigInteger.valueOf(0))){System.out.println("haha");}
 //            if(a.equals(zero)) {System.out.println("hehe");}
             c = a.add(b);
             if(cas > ) System.out.println();//大数的换行输出
             System.out.println("Case " + cas + ":");//大数的输出是用+号连接
             System.out.println(a + " + " + b + " = "+c);
         }
         cin.close();//关闭读入器
     }

 }

下面是大数string模拟的模板

 //***********加法*********************
 #include<algorithm>
 string add(string s1,string s2)
 {
       string ans = "";
       int i,j,x,y,k=;
       for(i=s1.length()-,j=s2.length()-;i>= && j>= ;i--,j--)
       {
          x = s1[i] - '';
          y = s2[j] - '';
          ans += char((x+y+k)% + '');
          k = (x+y+k)/;
       }
       while(i>=)
       {
          x=s1[i]-'';
          ans += char ((x+k)% + '');
          k = (x+k)/;
          i--;
       }
       while(j>=)
       {
          y=s2[j]-'';
          ans += char((y+k)% + '');
          k = (y+k)/;
          j--;
       }
       if(k>)
       ans += '';
       //ans.reverse();
       reverse(ans.begin(),ans.end());
       return ans;
 }
 //*******************************

 //************加法***************
 string add(string s1,string s2)
 {
       string ans = "";
       int i,j,x,y,k=;
       for(i=s1.length()-,j=s2.length()-;i>= && j>= ;i--,j--)
       {
          x = s1[i] - '';
          y = s2[j] - '';
          ans = char((x+y+k)% + '') + ans;
          k = (x+y+k)/;
       }
       while(i>=)
       {
          x=s1[i]-'';
          ans = char ((x+k)% + '') + ans;//不如+=快，但是可以不用倒序
          k = (x+k)/;
          i--;
       }
       while(j>=)
       {
          y=s2[j]-'';
          ans = char((y+k)% + '') + ans;
          k = (y+k)/;
          j--;
       }
       if(k>)
       ans = '' + ans;
       return ans;
 }
 //*********************加法**************************************

下面是完整的代码

 #include<cstdio>
 #include<iostream>
 #include<string>
 #include<cstring>
 #include<sstream>
 #include<algorithm>
 using namespace std;

 string add(string s1,string s2)
 {
       string ans = "";
       int i,j,x,y,k=;
       for(i=s1.length()-,j=s2.length()-;i>= && j>= ;i--,j--)
       {
          x = s1[i] - '';
          y = s2[j] - '';
          ans += char((x+y+k)% + '');
          k = (x+y+k)/;
       }
       while(i>=)
       {
          x=s1[i]-'';
          ans += char ((x+k)% + '');
          k = (x+k)/;
          i--;
       }
       while(j>=)
       {
          y=s2[j]-'';
          ans += char((y+k)% + '');
          k = (y+k)/;
          j--;
       }
       if(k>)
       ans += '';
       //ans.reverse();
       reverse(ans.begin(),ans.end());
       return ans;
 }
 int main()
 {
     string t , tt;
     int T ,c =  ;
     cin>>T;
     while(T--)
     {
         c++;
         cin>>t>>tt;
         string ans = add(t,tt);
         if(c!=) cout<<endl;
         cout<<"Case "<<c<<":"<<endl;
         cout<<t<<" + "<<tt<<" = "<<ans<<endl;
     }
     return ;
 }