UVa 465 溢出

来源：互联网发布：西方国家的虚伪知乎编辑：程序博客网时间：2024/06/15 15:43

* 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 300char 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;}

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