(HDU 1002)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<iostream>
#include<cstdio>
#include<algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include<sstream>
#include<set>
using namespace std;
int main()
{
int T,k=1;
cin>>T;
for(int p=1;p<=T;p++)
{
int x,y,num;
char a[1005],b[1005];
int c[1005],f=0;
memset(c,0,sizeof(c));
scanf("%s %s",a,b);
x=strlen(a);y=strlen(b);
printf("Case %d:\n%s + %s = ",k++,a,b);
num=0;x--;y--;
while(x>=0&&y>=0)
{
if(c[num]+a[x]-48+b[y]-48<10)
{c[num]+=a[x]-48+b[y]-48;}
else
{c[num]=c[num]+(a[x]-48+b[y]-48)-10;c[num+1]++;}
num++;
x--;
y--;
}
if(x>=0)
{
for(int t=x;t>=0;t--)
{
if(c[num]+a[t]-48<10)
{c[num]+=a[t]-48;}
else
{c[num]=(c[num]+a[t]-48)-10;c[num+1]++;}
num++;
}
}
else if(k>=0)
{
for(int t=y;t>=0;t--)
{
if(c[num]+b[t]-48<10)
{c[num]+=b[t]-48;}
else
{c[num]=(c[num]+b[t]-48)-10;c[num+1]++;}
num++;
}
}
else if(c[num]!=0)
num++;
for(int i=num;i>=0;i--)
{
if(c[i]==0&&f==0)
continue;
else
{
f=1;
printf("%d",c[i]);
}
}
printf("\n");
if(T!=p)
printf("\n");
}
return 0;
}