hdu1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 115102    Accepted Submission(s): 21803

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>

int a[1001], b[1001];
char charA[1001], charB[1001];
int sum[1001];

int main(){
    int n, i, j;
    scanf("%d", &n);
    for(i=1; i<=n; i++){
        if(i!=1){
            printf("\n");
        }
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        memset(sum, 0, sizeof(sum));
        printf("Case %d:\n", i);
        scanf("%s %s", &charA, &charB);
        printf("%s + %s = ", charA, charB);
        int lenA = strlen(charA);
        int lenB = strlen(charB);
        int p ,q;
        p=lenA;
        q=lenB;
        for(j=0; j<lenA; j++){
            a[p-1] = charA[j]-'0';
            p--;
        }
        for(j=0; j<lenB; j++){
            b[q-1] = charB[j]-'0';
            q--;
        }

        for(j=0; j<1001; j++){
            sum[j]+=a[j]+ b[j];
            sum[j+1]=sum[j]/10;
            sum[j]%=10;
        }
        
        for(j=1001; j>=0; j--){
              if(sum[j]!=0){
                  break;
              }
        }    
        int len = j;
        
        for(j=len; j>=0; j--){
            printf("%d", sum[j]);
        }
        printf("\n");
    }    
    
    return 0;
}