A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 166925 Accepted Submission(s): 31927
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>
#define MAX 1000
int main()
{
char an1[MAX+100];
char an2[MAX+100];
int an3[MAX+100];
int an4[MAX+100];
int n,k=0,p=0;
scanf("%d",&n);
while(n--)
{
int i,j,m,l;
scanf("%s%s",an1,an2);
memset(an3,0,sizeof(an3));
memset(an4,0,sizeof(an4));
l=strlen(an1);
for(i=l-1,j=0;i>=0;i--)
an3[j++]=an1[i]-'0';
m=strlen(an2);
for(i=m-1,j=0;i>=0;i--)
an4[j++]=an2[i]-'0';
for(i=0;i<MAX+100;i++)
{
an3[i]+=an4[i];
if(an3[i]>=10)
{
an3[i]-=10;
an3[i+1]+=1;
}
}
k++;
printf(p++?"
Case %d:
":"Case %d:
",k);
//printf("Case %d:
%s + %s = ",k,an1,an2);
printf("%s + %s = ",an1,an2);
for(i=MAX+99;(i>0)&&(an3[i]==0);i--);
for(j=i;j>=0;j--)
printf("%d",an3[j]);
printf("
");
}
return 0;
}