UVa 424 整数查询

/*

* 解题思路:

* 题意: 简单的大整数相加

*       几个注意点:1、注意输入两个数位数如果不同的位数控制

*                               2、最高位加完后可能存在向高一位的进位( 容易忽略 )

*/

#include <stdio.h>
#include <string.h>
#define A 120
int x,y,in;
int len1,len2;
char s[ A ];
char sum[ 2*A ];
void calc( int len , int pos )
{
    int i;
    if( !pos ) return;
    for( i=pos;i>=0;i-- )
    {
        x =     s[ i ] -‘0‘;
        y = sum[ i ] -‘0‘;

        if( x+y+in <= 9 &&  x+y+in >=0 )
        {
            sum[ i ] = ( char)( x+y+in+‘0‘ );
            in = 0;
        }
        else if( x+y+in > 9)
        {
            sum[ i ] = ( char )( (x+y+in)%10 )+‘0‘;
             in = 1;
        }
    }
}
int main( )
{
    int i;

    in = 0;
    memset( s,‘0‘,sizeof( s ) );
    scanf("%s",s );
    strcpy( sum , s );
    while( scanf("%s",s ) )
    {
        len1 = strlen( s );
        len2 = strlen( sum );
        if( len1 == 1 && s[ 0 ] == ‘0‘ ) break;

        if( len1 < len2 )
        {
           for(i=len1-1 ; i>=0 ;i-- )
                s[ i+len2-len1 ] = s[ i ];
            for( i=0;i<len2-len1;i++ )
                s[ i ] =‘0‘;
        }
        else if( len1 > len2 )
        {
            for( i=len2-1 ; i>=0;i-- )
                sum[ i+len1-len2 ] = sum[ i ];
            for( i=0;i<len1-len2;i++ )
                sum[ i ] = ‘0‘;
        }
        len1>=len2 ? calc( len1 , len1-1 ) : calc( len2 , len2-1 );
        if( in )
        {
            for( i=len2;i>0;i-- )
                sum[ i ] = sum[ i-1 ];
            sum[ 0 ] = ‘1‘;
            in = 0;
        }
    }

    printf("%s\n",sum);
    return 0;
}


UVa 424 整数查询

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