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

 

#include<cstdio>
#include<cstring>
void plus(int *a,int *b)
{
    int k;
    k=a[0]>b[0]?a[0]:b[0];
    for(int i=1; i<=k; i++)
    {
        a[i+1]+=(a[i]+b[i])/10;
        a[i]=(a[i]+b[i])%10;
    }
    if(a[k+1]) a[0]=k+1;
    else a[0]=k;
}
int main()
{
    char s1[1005],s2[1005];
    int a[1005],b[1005];
    int T,i,n,m,t;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {

        n=m=1;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        scanf("%s",s1);
        scanf("%s",s2);
        for(i=strlen(s1)-1; i>=0; i--)
            a[n++]=s1[i]-'0';
        a[0]=n-1;
        for(i=strlen(s2)-1; i>=0; i--)
            b[m++]=s2[i]-'0';
        b[0]=m-1;

        printf("Case %d:
",t);
        printf("%s + %s",s1,s2);
        printf(" = ");
        plus(a,b);
        for(i=a[0]; i>=1; i--)
            printf("%d",a[i]);
        printf("
");
        if(t!=T) printf("
");
    }
    return 0;
}
/*前几次一直PE，后来才发现原来是最后面多输出了一个回车，囧*/