hdoj1002A + B Problem II

发布时间：2020-12-14 02:21:03 所属栏目：大数据来源：网络整理

导读：A + B Problem II Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 266282????Accepted Submission(s): 51573 Problem Description I have a very simple problem for you. Given two integers

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 266282????Accepted Submission(s): 51573

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

??? 一看就是大数！！

? 代码：

#include<stdio.h>
#include<string.h>
char  al[1010],bl[1010];
int a[1010],b[1010];
int main()
{
    int i,j,n;
    scanf("%d",&n);
    getchar();
    int k=1;
     while(n--)
    {
        scanf("%s",al);//存为字符串 
        scanf("%s",bl);
        int l1=strlen(al);
        int l2=strlen(bl);
        memset(a,sizeof(a));//初始化 
        memset(b,sizeof(b));
        printf("Case %d:n",k++);
        for(j=0,i=l1-1;i>=0;i--)//转化为数字 
          {
              a[j]=al[i]-'0';
              j++;
          }
          for(j=0,i=l2-1;i>=0;i--)
           {
               b[j]=bl[i]-'0';
               j++;
           }
           for(i=0;i<1009;i++)//位位相加 
              {
                  a[i]=a[i]+b[i];
                  if(a[i]>=10)
                  {
                      a[i]-=10;
                     a[i+1]++;//如果大于十下一位加一 
                }
              }
              printf("%s + %s = ",al,bl);
              for(i=1009;i>=0;i--)
              {
                  if(a[i]!=0)//去掉前导零 
                    break;
              }
              if(i>=0)
               for(;i>=0;i--)
                 {
                     printf("%d",a[i]);
                 }
                 printf("n"); 
                 if(n!=0)
                 printf("n");
    }
}

（编辑：李大同）

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