hdu 1003 A + B Problem II 使用整型数组轻松实现大数求和

为题如下：

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
  
  

 

 

 

解体代码如下，实现很简单，这里就不说明实现过程

 

 

 
 

/********************************************** 
  
  
    > File Name: 1002.c 
  
  
    > Author: sea 
  
  
    > Mail: [email protected] 
  
  
    > Created Time: Mon 06 May 2013 02:01:19 PM CST 
  
  
 **********************************************/ 
  
  
 
  
  
#include<stdio.h> 
  
  
#include<string.h> 
  
  
 
  
  
#define N 1101 
  
  
 
  
  
int main() { 
  
  
    char str1[N],str2[N]; 
  
  
    int sum[N+1],i,j,m,n,t,s1,s2,k,flag=1; 
  
  
    scanf("%d",&i); 
  
  
    while(i--) { 
  
  
        k=N-1; 
  
  
        memset(str1,sizeof(str1)); 
  
  
        memset(str2,sizeof(str2)); 
  
  
        memset(sum,sizeof(sum)); 
  
  
        scanf("%s %s",str1,str2); 
  
  
        m=strlen(str1)-1; 
  
  
        n=strlen(str2)-1; 
  
  
        if(m>n) t=m; 
  
  
        else t=n; 
  
  
        for(j=0;j<=t;j++,m--,n--,k--) { 
  
  
            if(m<0) s1=0; 
  
  
            else s1=str1[m]-48; 
  
  
            if(n<0) s2=0; 
  
  
            else s2=str2[n]-48; 
  
  
            if(sum[k]+s1+s2>=10) 
  
  
                sum[k-1]+=1; 
  
  
            sum[k]=(sum[k]+s1+s2)%10; 
  
  
 
  
  
        } 
  
  
        sum[k]>0?k:k++; 
  
  
        printf("Case %d:\n%s + %s = ",flag++,str2); 
  
  
        for(j=k;j<N;j++) 
  
  
            printf("%d",sum[j]); 
  
  
        if(i>=1)    printf("\n\n"); 
  
  
        else printf("\n"); 
  
  
    } 
  
  
    return 0; 
  
  
}