HDU 1002 A + B Problem II 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): 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

21 2112233445566778899 998877665544332211

Sample Output

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

/*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 = ",a,b);            for(i=k-1;i>=0;i--)            {                cout<<c[i];            }            printf("\n");            if(num<=n)     printf("\n");        }         return 0;}