HDU 1002 A + B Problem II 大数相加

java  水过

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

public class aa
{
	public static void main( String args[] ) 
	{
		BigInteger a,b;
		Scanner in = new Scanner ( System.in );
		int t = in.nextInt();
		for( int i = 1; i <= t; ++i )
		{
			in.hasNext();
			a = in.nextBigInteger();b = in.nextBigInteger();
			BigInteger c = a.add(b);
			System.out.println( "Case "+i+":" );
			System.out.println( a+" + "+b+" = "+c );
			if( i != t )
				System.out.println(  );
		}
	}
}

刚刚开始系统的做大数,又把这题水了一遍,不过太大意的,输出格式错了几次,下次就算是做现题都要认真;

因为"大数"可能很大,所以long long和double可能都存不下,所以要用数组来把他们存下来,然后一位一位来计算;而且要从个位数即数组的最后一位加起

#include<stdio.h>
#include<string.h>
const int Maxsize = 1024;
char num[Maxsize],a[Maxsize];
int sum[Maxsize];
int main( )
{
    int t,k;
    scanf( "%d\n",&t );
    k = t;
    while( k-- )
    {
           memset( sum,0,sizeof( sum ) );
           for( int i = 0; i < 2; ++i )
           {
                scanf( "%s",num );
                int len = strlen( num );
                for( int i = 0; i < len; ++i )//核心
                {
                     sum[i] += num[len - i - 1] - '0';
                     if( sum[i] > 9 )
                     {
                         ++sum[i+1];
                         sum[i] -= 10;
                         }
                     
                }
                if( i == 0 )
                    strcpy( a,num );
           }
           int j = Maxsize - 1;
           while( !sum[j] )
                  --j;
           printf( "Case %d:\n",t - k );
           printf( "%s + %s = ",a,num );
           for( ; j >= 0; --j )
                printf( "%d",sum[j] );
           puts( "" );
           if( k )puts( "" );
           } 
    return 0;
}