HDU水题 A+BII水题

</pre>A + B Problem II</h1><span size="+0" style=""><strong><span style="font-size:12px; font-family:Arial; color:green">Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 261678    Accepted Submission(s): 50648</span></strong></span><div class="panel_title" align="left">Problem Description</div><div class="panel_content">I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.</div><div class="panel_bottom"> </div><div class="panel_title" align="left">Input</div><div class="panel_content">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.</div><div class="panel_bottom"> </div><div class="panel_title" align="left">Output</div><div class="panel_content">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.</div><div class="panel_bottom"> </div><div class="panel_title" align="left">Sample Input</div><div class="panel_content"><pre><div style="font-family:Courier New,Courier,monospace">21 2112233445566778899 998877665544332211</div>

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

 //解题思路：这道题快把折磨疯了，注意输出问题，注意大数相加，我用的是模板，参考代码如下

#include<stdio.h>#include<string.h>char a[2000],b[2000];int  A[2000],B[2000];int main(){ int n,x; scanf("%d",&n); getchar(); for(x=1;x<=n;x++)  {  int i,j,k,w;    printf("Case %d:\n",x);        scanf("%s%s",a,b);           k=strlen(a);           w=strlen(b);           memset(A,0,sizeof(A));           memset(B,0,sizeof(B));        printf("%s + %s = ",a,b); for(j=0,i=k-1;i>=0;i--)   {  A[j++]=a[i]-'0';   }  for(j=0,i=w-1;i>=0;i--)   { B[j++]=b[i]-'0';     }     for(i=0;i<2000;i++)     {  A[i]+=B[i];       if(A[i]>=10)          { A[i]-=10;            A[i+1]++;          }             }     for(i=1999;i>=0&&A[i]==0;i--);     for(;i>=0;i--)       {printf("%d",A[i]);       }       printf("\n");      if(x!=n)       {printf("\n");       }  

    }return 0;}