/*
* 解题思路：
* 题意很好理解、题目也没有陷阱、但是如果想用通俗方法( 不是网上很多就简单几行代码过题写法 )写这道题还是有些难度的、注意点如下
* 1、边界问题、上界是2147483647、自己可以测几组边界数据监测（ 2147483647+1 和 2147483647 * 1 ）
* 2、内部测试数据可以定有前缀为0的数、这里一定要判断好( 00000000000001 * 00000000000006 )
* 3、第三点我错了很久、先前直接只要两个数、任何一个数溢出、则我判断答案溢出，但是忘记考虑任何数乘以0都为0了!!
*/
#include <stdio.h>
#include <string.h>
#define A 300
char sample[ 15 ] = "2147483647";
char ss1[ A ],ss2[ A ],c[ 5 ];
char x[ A ];
int in;
int Judge( char sss[] )
{
int i;
for( i=0;i<10;i++ )
{
if( sss[ i ]-‘0‘> sample[ i ]-‘0‘ ) return 1;
if( sss[ i ]-‘0‘ < sample[ i ]-‘0‘ ) return 0;
}
return 0;
}
void calc( int len , int pos ,char s[] , char sum[] )
{
int i;
int x,y;
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 Mutiply( char s1[] ,char s2[] , int len1 ,int len2 )
{
int x,y,z,zz=0;
int i,j,k;
int tmp,flag;
int len;
char ss[ A ],s[ A ];
if( len1 > len2 )
{
strcpy( s , s1 );
strcpy( s1,s2 );
strcpy( s2 , s );
tmp = len1;
len1 = len2;
len2 = tmp;
}
memset( ss , ‘0‘ , sizeof( ss ) );
for( i=len2-1;i>=0;i-- )
{
in = 0;
memset( s , ‘0‘, sizeof( s ) );
for( j=len1-1;j>=0;j-- )
{
x = s1[ j ] - ‘0‘;
y = s2[ i ] - ‘0‘;
if( ( x*y+in) >= 10 )
{
s[ i+j ] = ( char )(( (x*y+in)%10 )+‘0‘);
in = (x*y+in)/10;
}
else if( (x*y+in )<=9 && (x*y+in ) >=0 )
{
s[ i+j ] =(char) (x*y+in +‘0‘);
in = 0;
}
}
if( in != 0 && i!=0 )
s[ i-1 ] = in+‘0‘;
flag = 0;
for( j=i+len2-1;j>=0;j-- )
{
if( ss[ j ]-‘0‘ + s[ j ] -‘0‘ + flag >=10 )
{
ss[ j ] = (ss[ j ]-‘0‘ + s[ j ] - ‘0‘ + flag )%10+‘0‘;
flag = 1;
}
else
{
ss[ j ] += s[ j ] -‘0‘+ flag;
flag = 0;
}
}
if( i == 0 && in+flag )
{
zz = 0;
memset( s ,‘\0‘, sizeof( s ) );
z = in + flag;
while( z!=0 )
{
s[ zz++ ] = z%10+‘0‘;
z/=10;
}
ss[ len1+len2-1+zz ] = ‘\0‘;
len = strlen( ss );
for( k=len+zz;k>=zz;k-- )
ss[ k ] = ss[ k-zz ];
for( k=0;k<zz;k++ )
ss[ zz-k-1 ] = s[ k ];
}
}
ss[ len1+len2-1+zz ] = ‘\0‘;
len = strlen( ss );
if( len >10 ) return 1;
else if( len <10 ) return 0;
else if( len == 10 ) return Judge( ss );
}
int Plus( char s[] ,char sum[] , int len1 , int len2 )
{
int i;
in = 0;
if( len1 == 1 && s[ 0 ] == ‘0‘ ) return 0;
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‘;
sum[ len2 ] = ‘\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‘;
sum[ len1 ] = ‘\0‘;
}
len1>=len2 ? calc( len1 , len1-1 , s , sum ) : calc( len2 , len2-1 , s , sum );
if( in )
{
for( i=len2;i>0;i-- )
sum[ i ] = sum[ i-1 ];
sum[ 0 ] = ‘1‘;
in = 0;
}
len2 = strlen( sum );
if( len2 > 10 ) return 1;
else if( len2 < 10 ) return 0;
else if( len2 == 10 ) return Judge( sum );
}
int main( )
{
int i,j;
int flag,vis,sum;
int len1,len2;
char str[ 5 ][ 30 ] ={"first number too big","second number too big","result too big"};
memset( ss1 , ‘0‘ , sizeof( ss1 ) );
memset( ss2 , ‘0‘ , sizeof( ss2 ) ) ;
while( ~scanf("%s%s%s",ss1,c,ss2) )
{
flag = vis = sum = 0;
len1 = strlen( ss1 );
len2 = strlen( ss2 );
printf("%s %s %s\n",ss1,c,ss2 );
for( i=0;i<len1;i++ )
if(!vis && ss1[ i ] == ‘0‘ ) sum++;
else break;
if( sum )
{
j = 0;
for( i=sum;i<len1;i++ )
ss1[ j++ ] = ss1[ i ];
len1 = j;
ss1[ j ] = ‘\0‘;
}
vis = sum =0;
for( i=0;i<len2;i++ )
if(!vis && ss2[ i ] == ‘0‘ ) sum++;
else break;
if( sum )
{
j = 0;
for( i=sum;i<len2;i++ )
ss2[ j++ ] = ss2[ i ];
len2 = j;
ss2[ j ] = ‘\0‘;
}
if( len1 > 10 || ( len1 == 10 && Judge( ss1 ) ) )
printf("%s\n",str[ 0 ] );
if( len2 > 10 || ( len2 ==10 && Judge( ss2 ) ) )
printf("%s\n",str[ 1 ] );
if( c[ 0 ] == ‘+‘ && len1!=0 && len2 !=0 )
if( Plus( ss1 , ss2 , len1 , len2 ) ) flag = 1;
if( c[ 0 ] == ‘*‘ && len1!=0 && len2 !=0 )
if( Mutiply( ss1 , ss2 , len1 , len2 ) ) flag = 1;
if( flag ) printf("%s\n",str[ 2 ] );
}
return 0;
}