hdu1002 A + B Problem II(大数相加)

发布时间：2020-12-14 03:42:29 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): 189701????Accepted Submission(s): 36249 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): 189701????Accepted Submission(s): 36249

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

#include<stdio.h>
#include<string.h>
char s1[1004],s2[1004];
int a[1004],b[1004];
int main()
{
    int t,u,i,j,c;
    scanf("%d",&t);
    for(u=1;u<=t;u++)
    {
        memset(a,sizeof(a));
        memset(b,sizeof(b));
        scanf("%s%s",s1,s2);
        if(u!=1) printf("n");
        printf("Case %d:n%s + %s = ",s2);
        j=0;
        for(i=strlen(s1)-1;i>=0;i--)
            a[j++]=s1[i]-'0';
        j=0;
        for(i=strlen(s2)-1;i>=0;i--)
            b[j++]=s2[i]-'0';
        i=c=0;
        while(i<strlen(s1)||i<strlen(s2)||c)
        {
            a[i]+=b[i]+c;
            c=0;
            if(a[i]>9)
            {
                a[i]-=10;
                c=1;
            }
            i++;
        }
        for(--i;i>=0;i--)
            printf("%d",a[i]);
        printf("n");
    }
    return 0;
}

（编辑：李大同）

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