A+B Problem II（高精度运算）

A+B Problem II

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

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

A,B must be positive.

输入

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.

输出

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.

样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

c语言实现：

02.#include<iostream> 
03.#include<cstdio> 
04.#include<cstring>  
05.#include<cstdlib>  
06.#define N 1001 
07.using namespace std;
08.    
09.char ch1[N],ch2[N]; 
10.int num1[N],num2[N]; 
11.int main() 
12.{ 
13.  int count,i,j,len1,len2,len,T; 
14.  scanf("%d",&T); 
15.  count=0; 
16.     
17.   do 
18.   { 
19.    ++count; 
20.    scanf("%s%s",ch1,ch2);            
21.    len1=strlen(ch1); 
22.    len2=strlen(ch2);
23.    memset(num1,0,N*sizeof(num1[0])); 
24.    memset(num2,0,N*sizeof(num2[0])); 
25.      
26.    for(i=len1-1,j=0;i>=0;--i,++j)//从个位开始转化并存放 
27.     num1[j]=ch1[i]-'0';
28.       
29.    for(i=len2-1,j=0;i>=0;--i,++j)//从个位开始转化并存放 
30.     num2[j]=ch2[i]-'0';
31.       
32.    len =len1>len2 ? len1:len2;
33.    for(i=0;i<len;++i)
34.    {
35.       num1[i]+=num2[i];
36.       if(num1[i]>9&&i<len)
37.       {
38.         num1[i]-=10;
39.         num1[i+1]++;//进位 
40.       }                        
41.    }
42.      
43.      printf("Case %d:\n",count); 
44.      printf("%s + %s = ",ch1,ch2);
45.        
46.       if(num1[len]) printf("%d",num1[len]);//可能存在的进位 
47.       for(i=len-1 ; i >=0; --i) 
48.       printf("%d",num1[i]); 
49.       printf("\n");  
50.   }while(--T);  
51.     
52. //   system("pause"); 
53.    return 0;   
54.} 

JAVA实现：

02.import java.math.BigInteger;
03.import java.util.Scanner;
04.public class Main{
05.  
06.    public static void main(String args[]) {
07.       Scanner cin=new Scanner(System.in);
08.       int n=cin.nextInt();
09.       BigInteger a,b;
10.       for(int i=1;i<=n;i++){
11.           a=cin.nextBigInteger();
12.           b=cin.nextBigInteger();
13.           System.out.println("Case "+i+":");
14.           System.out.println(a.toString()+" + "+b.toString()+" = "+a.add(b));
15.       }
16.    }
17.}