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

 1 import java.math.BigInteger;

 2 import java.util.Scanner;

 3 

 4 

 5 public class Main {

 6     public static void main(String[] args) {

 7         Scanner cin = new Scanner(System.in);//大数的输入，定义一个输入器

 8         BigInteger a = null, b = null, c = null;//开始要赋值成空

 9         a = BigInteger.valueOf(100);

10         b = BigInteger.valueOf(99);

11         int T;

12         T = cin.nextInt();//读入T;

13 //        while(cin.hasNextBigInteger())//判断是否读到文件结尾相当于while(~scanf())

14         for(int cas = 1; cas <= T; cas++)

15         {

16             a = cin.nextBigInteger();

17             b = cin.nextBigInteger();

18             

19 //            BigInteger zero = BigInteger.valueOf(0);//大数判断是不是等于0

20 //            if(a.equals(BigInteger.valueOf(0))){System.out.println("haha");}

21 //            if(a.equals(zero)) {System.out.println("hehe");}

22             c = a.add(b);

23             if(cas > 1) System.out.println();//大数的换行输出

24             System.out.println("Case " + cas + ":");//大数的输出是用+号连接

25             System.out.println(a + " + " + b + " = "+c);

26         }

27         cin.close();//关闭读入器

28     }

29 

30 }

 

下面是大数string模拟的模板

 1 //***********加法*********************

 2 #include<algorithm>

 3 string add(string s1,string s2)      

 4 {

 5       string ans = "";

 6       int i,j,x,y,k=0;

 7       for(i=s1.length()-1,j=s2.length()-1;i>=0 && j>=0 ;i--,j--)

 8       {

 9          x = s1[i] - '0';

10          y = s2[j] - '0';

11          ans += char((x+y+k)%10 + '0');

12          k = (x+y+k)/10;

13       }

14       while(i>=0)

15       {

16          x=s1[i]-'0';

17          ans += char ((x+k)%10 + '0');

18          k = (x+k)/10;

19          i--;

20       }

21       while(j>=0)

22       {

23          y=s2[j]-'0';

24          ans += char((y+k)%10 + '0');

25          k = (y+k)/10;

26          j--;

27       }

28       if(k>0)

29       ans += '1';

30       //ans.reverse();

31       reverse(ans.begin(),ans.end());

32       return ans;

33 }

34 //*******************************

35 

36 //************加法***************

37 string add(string s1,string s2)      

38 {

39       string ans = "";

40       int i,j,x,y,k=0;

41       for(i=s1.length()-1,j=s2.length()-1;i>=0 && j>=0 ;i--,j--)

42       {

43          x = s1[i] - '0';

44          y = s2[j] - '0';

45          ans = char((x+y+k)%10 + '0') + ans;

46          k = (x+y+k)/10;

47       }

48       while(i>=0)

49       {

50          x=s1[i]-'0';

51          ans = char ((x+k)%10 + '0') + ans;//不如+=快，但是可以不用倒序

52          k = (x+k)/10;

53          i--;

54       }

55       while(j>=0)

56       {

57          y=s2[j]-'0';

58          ans = char((y+k)%10 + '0') + ans;

59          k = (y+k)/10;

60          j--;

61       }

62       if(k>0)

63       ans = '1' + ans;

64       return ans;

65 }

66 //*********************加法**************************************

 

下面是完整的代码

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<string>

 4 #include<cstring>

 5 #include<sstream>

 6 #include<algorithm>

 7 using namespace std;

 8 

 9 string add(string s1,string s2)

10 {

11       string ans = "";

12       int i,j,x,y,k=0;

13       for(i=s1.length()-1,j=s2.length()-1;i>=0 && j>=0 ;i--,j--)

14       {

15          x = s1[i] - '0';

16          y = s2[j] - '0';

17          ans += char((x+y+k)%10 + '0');

18          k = (x+y+k)/10;

19       }

20       while(i>=0)

21       {

22          x=s1[i]-'0';

23          ans += char ((x+k)%10 + '0');

24          k = (x+k)/10;

25          i--;

26       }

27       while(j>=0)

28       {

29          y=s2[j]-'0';

30          ans += char((y+k)%10 + '0');

31          k = (y+k)/10;

32          j--;

33       }

34       if(k>0)

35       ans += '1';

36       //ans.reverse();

37       reverse(ans.begin(),ans.end());

38       return ans;

39 }

40 int main()

41 {

42     string t , tt;

43     int T ,c = 0 ;

44     cin>>T;

45     while(T--)

46     {

47         c++;

48         cin>>t>>tt;

49         string ans = add(t,tt);

50         if(c!=1) cout<<endl;

51         cout<<"Case "<<c<<":"<<endl;

52         cout<<t<<" + "<<tt<<" = "<<ans<<endl;

53     }

54     return 0;

55 }