hdoj1002--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

 

 

簡單題：大數的運算

注意格式（Case的首字母大寫、各類空格、每一行之間有空行，最後一行沒有空行）！

給出幾組測試數據:

6

1 2

1 0

9999 1

1 999999

5555 4445

112233445566778899 998877665544332211

 

java code:

import java.util.*;
import java.io.*;
import java.math.*;

class Main
{
    public static BigInteger plus(BigInteger a, BigInteger b) {
        BigInteger c;
        c = a.add(b);
        return c;
    }
    
    public static void main(String args[])
    {
        Scanner cin = new Scanner(System.in);
        int T, i;
        BigInteger a,b;
        T = cin.nextInt();
        i = 1;
        while((T--) > 0) {
            a = cin.nextBigInteger();
            b = cin.nextBigInteger();
            System.out.println("Case "+ i +":");
            System.out.println(a + " + "+ b + " = "+ plus(a, b));
            i++;
            if(T > 0)
                System.out.println();
        }
    }
}

C code

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
void solve()
{
    int n,i,j,k,flag,t,cas,L;
    char a[1001],b[1001],c[1002];
    scanf("%d",&n);
    getchar();
    cas=1;
    while(n--)
    {
        flag=0;
        memset(a,'\0',sizeof(a));
        memset(b,'\0',sizeof(b));
        memset(c,'\0',sizeof(c));
        scanf("%s",a);
        scanf("%s",b);
        printf("Case %d:\n",cas);
        printf("%s",a);
        printf(" + ");
        printf("%s",b);
        printf(" = ");
        strrev(a);
        strrev(b);
        k=i=0;
        L=(strlen(a)>strlen(b)?strlen(b):strlen(a));
        while(i<L)
        {
            t=(a[i]-'0')+(b[i]-'0')+flag;
            flag=(t>=10?1:0);
            c[k++]=t%10+'0';
            i++;
        }
        if(a[i]=='\0')
        {
            while(b[i]!='\0') {
                t=b[i++]-'0'+flag;
                c[k++]=t%10+'0';
                flag=(t>=10?1:0);
            }
        }
        else
        {
            while(a[i]!='\0') {
                t=a[i++]-'0'+flag;
                c[k++]=t%10+'0';
                flag=(t>=10?1:0);
            }
        }
        if(flag) c[k]='1';
        else k--;
        while(k>=0)
        {
            printf("%c",c[k]);
            k--;
        }
        printf("\n");
        if(n>0) printf("\n");
        cas++;
    }
}

int main()
{
    solve();
    return 0;
}