# HDOJ1002题A + B Problem II，2个大数相加

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 < stdio.h>
#include < stdlib.h>
#include < string.h>
int main()
{
char a[1010],b[1010],c[1010];
int a1,b1,m,i,l,i1,j1,n1,m1,a2,b2,j=1,t1,t2,n,p=0;
scanf("%d",&n);
while(n--)
{
p=0;
scanf("%s",a);
scanf("%s",b);
printf("Case %d:\n",j++);
printf("%s + %s = ",a,b);
a1=strlen(a);
b1=strlen(b);
a2=a1;
b2=b1;
for(i=0; a1>=0||b1>=0; i++,a1--,b1--)
{
if(a1>=0&&b1>=0)
{
c[i]=a[a1]+b[b1]-'0'+p;
}
else if(a1>=0&&b1<0)
{
c[i]=a[a1]+p;
}
else if(a1<0&&b1>=0)
{
c[i]=b[b1]+p ;
}
p=0;
if(c[i]>'9')
{
c[i]=c[i]-10;
p=1;
}
}
if(p==1)
printf("%d",p);
t1=1;
t2=i-1;
n1=m1=0;
for(i1=0; i1
{
if(a[i1]=='0')
n1++;
}
for(j1=0 ; j1
{
if(b[j1]=='0')
m1++;
}
if(n1==a2&&m1==b2)
{
printf("0");
}
else
{
for(l= i-1 ; l>0; l--)
{
if(t2==l&&c[l]=='0'&&p!=1)
{
t2--;
continue;
}
printf("%c",c[l]);
}
}
if(n!=0)
printf("\n\n");
else
printf("\n");
}
return 0;
}
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