HDU 1002 A + B Problem II 大数相加

发布时间：2020-12-14 03:31:51 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): 204234????Accepted Submission(s): 39250 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): 204234????Accepted Submission(s): 39250

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

/*
1002 A + B Problem II
大数加法？ 
*/
#include<iostream>
#include<cstring>
using namespace std;

char a[1001],b[1001];
int c[1001];

int main(){
    int i,j,n,k,z,len1,len2,t,num; 
    
        
        cin>>n;
        num=1; 
        while(num<=n)
        {
            scanf("%s %s",a,b);
            len1=strlen(a)-1;
            len2=strlen(b)-1;
            k=0;
            z=0;
            for(;len1>=0&&len2>=0;len1--,len2--)
            {
            
                t=(a[len1]-'0')+(b[len2]-'0')+z;
                c[k++]=t%10;
                z=t/10;
            }
            while(len1>=0)
            {
                t=a[len1]-'0'+z;
                c[k++]=t%10;
                z=t/10;
                len1--;    
            }
            while(len2>=0)
            {
                t=b[len2]-'0'+z;
                c[k++]=t%10;
                z=t/10;
                len2--;    
            }
            printf("Case %d:n",num++);
            printf("%s + %s = ",b);
            for(i=k-1;i>=0;i--)
            {
                cout<<c[i];
            }
            printf("n");
            if(num<=n)     printf("n");
        }     
    return 0;
}

（编辑：李大同）

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