HDU 1002 大数A+B

　　　　　　　　　　　　　　A + B Problem II

 

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

 

代码：

#include<cstring>
#include<cstdio>
using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);
    int count,temp;
    int a1[1010],a2[1010];
    char s1[1010],s2[1010];
    int t,i,j,len1,len2;
    scanf("%d",&count);
    t=1;
    while(count--)
    {
        printf("Case %d:\n",t++);
        memset(a1,0,sizeof(a1));
        memset(a2,0,sizeof(a2));
        scanf("%s%s",s1,s2);
        printf("%s + %s = ",s1,s2);
        int len1=strlen(s1);
        int len2=strlen(s2);
        for(i=len1-1,j=0;i>=0;i--)
        a1[j++]=s1[i]-'0';
        for(i=len2-1,j=0;i>=0;i--)
        a2[j++]=s2[i]-'0';
        int maxlen=len1>len2?len1:len2;
        for(i=0;i<maxlen;i++)
        {
            a1[i]+=a2[i];
            if(a1[i]>=10)
            {
                a1[i+1]+=1;
                a1[i]-=10;
            }
        }
        for(j=1009;j>=0;j--)
        if(a1[j])break;
        if(j==-1)
        printf("0");
        for(;j>=0;j--)
        printf("%d",a1[j]);
        printf("\n");
        if(count!=0)
        printf("\n");
    }
    return 0;
}