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hdu 1002 A + B Problem II


A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209836    Accepted Submission(s): 40373




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


/*题解:
    发现杭电和南阳上的大数输出问题不同 ,今天又用java水了一份,java解大数真的很简单。 
    */

JAVA版:
import java.util.*;
import java.math.*;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in=new Scanner (System.in);
int i,n=in.nextInt();
BigInteger a,b,c;
for(i=1; i<=n; i++){
a=in.nextBigInteger();
b=in.nextBigInteger();
c=a.add(b);
System.out.println("Case "+i+":");
System.out.println(a+" + "+b+" = "+c);
if(i!=n) System.out.println("");              //print("\n");就会PE
}
}
}


C版:

#include<cstdio>


#include<cstring>

int main()

{

    int n,i,j;

    scanf("%d",&n);

    char str1[1010],str2[1010];

    int len1,len2;

    for(j=1; j<=n; j++)

    {

        scanf("%s %s",str1,str2);

        len1=strlen(str1);

        len2=strlen(str2);

       int num1[1010]={0},num2[1010]={0};

        /*memset(num1,0,sizeof(num1));

        memset(num2,0,sizeof(num2));}*/

        //反转并赋值 

        for(i=0; i<len1; i++)

        num1[len1-1-i]=str1[i]-'0';

        for(i=0; i<len2; i++)

        num2[len2-1-i]=str2[i]-'0';

        //令len1为最大 

        if(len1<len2)

        {

            int t=len1;

            len1=len2;

            len2=t;

        }

        for(i=0; i<=len1; i++)

        {

            num1[i] += num2[i];

            if(num1[i]>=10)

            {

                num1[i]-=10;

                num1[i+1]++;

            }

        }

        printf("Case %d:\n",j);


        printf("%s + %s = ",str1,str2);

         //去零


        if(num1[len1]==0)

        {

            for(i=len1-1; i>=0; i--)

            printf("%d",num1[i]);

            printf("\n");

        }

        if(num1[len1]!=0)

        {

            for(i=len1; i>=0; i--)

            printf("%d",num1[i]);

            printf("\n");

        }

        if(j!=n) printf("\n"); 

     }



    return 0;

}

    

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