A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 295229 Accepted Submission(s): 56839

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

大整数加法：字符数组接收，逆序成整形，然后进位相加，前导去掉，再逆序输出 

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int main(){
    int T;
    char a[1005],b[1005];
    int  x[1005],y[1005];
    int  ans[1005]; 
    scanf("%d",&T);
    for(int k=1;k<=T;k++){
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(ans,0,sizeof(ans));


        scanf("%s %s",a,b);
        int len1=strlen(a);
        int len2=strlen(b);
        for(int i=0;i<len1;i++)
            x[i]=a[len1-1-i]-'0';
        for(int i=0;i<len2;i++)
            y[i]=b[len2-1-i]-'0';        //逆序两数 
        int c=0;                        //代表进位 
        for(int i=0;i<len1+1||i<len2+1;i++){
            ans[i]=(x[i]+y[i]+c)%10;
            c=(x[i]+y[i]+c)/10;
        }                                //两者之和可能比原来的位数多1 

        printf("Case %d:\n",k);
        printf("%s + %s = ",a,b);

        int len=len1 > len2? len1 :len2;
        for(int i=len; i>=0;i--){
            if(i==len&&ans[i]==0)
             continue;
            else 
               {
                printf("%d",ans[i]);
               }
        }
        printf("\n");
        if(k!=T)
         printf("\n");  
    }

    return 0;
}