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

发布时间：2020-12-14 04:10:07 所属栏目：大数据来源：网络整理

导读：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 cas

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 = 2222222222222221110

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:?windorman@gmail.com?
  
  ????>?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("nn");?
  
  ????????else?printf("n");?
  
  ????}?
  
  ????return?0;?
  
  }?

（编辑：李大同）

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