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<stdio.h>
#include<string.h>
char a[1010],b[1010];
int sum[1010];
int main()
{
int i,j,k,s,lena,lenb,t,d;
while(scanf("%d",&t)!=EOF)
{
d=t;
s=0;
while(t--)
{
s++;
scanf("%s%s",a,b);
lena=strlen(a);
lenb=strlen(b);
memset(sum,0,sizeof(sum));
k=0;
for(i=lena-1,j=lenb-1;i>=0&&j>=0;i--,j--)
{
sum[k]=a[i]-'0'+b[j]-'0'+sum[k];
if(sum[k]>=10)
{
sum[k]=sum[k]-10;
sum[k+1]=sum[k+1]+1;
}
k++;
}
while(j>=0)
{
sum[k]=b[j--]-'0'+sum[k];
if(sum[k]>=10)
{
sum[k]=sum[k]-10;
sum[k+1]=sum[k+1]+1;
}
k++;
}
while(i>=0)
{
sum[k]=a[i--]-'0'+sum[k];
if(sum[k]>=10)
{
sum[k]=sum[k]-10;
sum[k+1]=sum[k+1]+1;
}
k++;
}
k=1009;
while(k--)
{
if(sum[k]!=0)
break;
}
printf("Case %d:\n",s);
printf("%s + %s = ",a,b);
if(k==-1)
printf("0");
for(i=k;i>=0;i--)
printf("%d",sum[i]);
printf("\n");
if(s!=d)
printf("\n");
}
}
return 0;
}