HDU 1002.A + B Problem II-数组模拟-大数相加

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 435932    Accepted Submission(s): 84825

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 = 1111111111111111110

 

Author

Ignatius.L

 

代码:

 #include<stdio.h>
 #include<string.h>
 int main(){
     char a[],b[];
     int c[],d[],ans[];
     int n,i,k,j,h,t,l;
     int len1,len2;
     scanf("%d",&n);
         h=;
         while(n--){
             h+=;
             for(i=;i<;i++){
                 c[i]=d[i]=;
                 ans[i]=;
             }
             scanf("%s %s",&a,&b);
             len1=strlen(a);
             len2=strlen(b);
             for(i=;i<len1;i++){
                c[i]=a[i]-'';
             }
             for(i=;i<len2;i++){
                d[i]=b[i]-'';
             }
             printf("Case %d:\n",h);
             for(i=;i<len1;i++)
                 printf("%d",c[i]);
                 printf(" + ");
             for(i=;i<len2;i++)
                 printf("%d",d[i]);
                 printf(" = ");
             for(i=,j=len1-;i<len1/;i++,j--){
                 t=c[j];
                 c[j]=c[i];
                 c[i]=t;
             }
             for(i=,j=len2-;i<len2/;i++,j--){
                 t=d[j];
                 d[j]=d[i];
                 d[i]=t;
             }
             l=len1>len2?len1:len2;
             for(i=,k=;i<=l;i++,k++){
                 if(c[i]+d[i]<)
                     ans[k]=c[i]+d[i];
                 else{
                     ans[k]=(c[i]+d[i])%;
                     c[i+]+=(c[i]+d[i])/;
                 }

             }
             for(j=k-;j>=;j--){
                 if(j==k-&&ans[j]==)
                     continue;
                 printf("%d",ans[j]);
             }
                 if(n==)
                     printf("\n");
                 else
                     printf("\n\n");
         }
     return ;
 }