大数除法 10494 - If We Were a Child Again

Problem C
If We Were a Child Again

Input: standard input
Output: standard output

Time Limit: 7 seconds

 

“Oooooooooooooooh!

If I could do the easy mathematics like my school days!!

I can guarantee, that I’d not make any mistake this time!!”

Says a smart university student!!

But his teacher even smarter – “Ok! I’d assign you such projects in your software lab. Don’t be so sad.”

“Really!!” - the students feels happy. And he feels so happy that he cannot see the smile in his teacher’s face.

 

 

The Problem

 

The first project for the poor student was to make a calculator that can just perform the basic arithmetic operations.

 

But like many other university students he doesn’t like to do any project by himself. He just wants to collect programs from here and there. As you are a friend of him, he asks you to write the program. But, you are also intelligent enough to tackle this kind of people. You agreed to write only the (integer) division and mod (% in C/C++) operations for him.

 

Input

Input is a sequence of lines. Each line will contain an input number. One or more spaces. A sign (division or mod). Again spaces. And another input number. Both the input numbers are non-negative integer. The first one may be arbitrarily long. The second number n will be in the range (0 < n < 2³¹).

 

 

Output

A line for each input, each containing an integer. See the sample input and output. Output should not contain any extra space.

 

 

 

Sample Input

110 / 100

99 % 10

2147483647 / 2147483647

2147483646 % 2147483647

 

 

 

 

 

 

Sample Output

1

9

1

2147483646

 

第一次做大数除法，其实和大数乘法很类似，模拟草稿纸上的运算。

有几个细节需要注意一下，第一个是在除数变化的过程中，可能会超过int型，所以用了double转换了一下

而且数组开得大一些，不然会RE

还有就是要注意到除数为0的特殊情况。一开始没有注意到，wa了几次。。

#include<iostream>#include<cstring>#include<cstdio>#include<ctype.h>#include<cmath>#include<algorithm>char big[1000000];char str[1000000],ch;int di[1000000], mod[1000000];int num;using namespace std;int main (){    int i,j,len;    while(gets(str)!=NULL)    {        memset(big,0,sizeof(big));        memset(di,0,sizeof(di));        memset(mod,0,sizeof(mod));        num=0;        len=strlen(str);        for (i=0; ; i++)            if (str[i]!=' ') big[i]=str[i];            else break;        for (j=i; ; j++)            if (str[j]!=' ')            {                ch=str[j];                break;            }        j++;        while(str[j]==' ') j++;        for (i=j; i<len; i++)            num=num*10+str[i]-'0';        for (i=0; i<strlen(big); i++)        {            double g,t;            t=mod[i];            g=t*10+big[i]-'0';            di[i]=(int)(g/num);            t=(int)(g/num);            mod[i+1]=g-t*num;        }        if (ch=='%') cout<<mod[i]<<endl;        else        {            for (i=0; i<strlen(big); i++)                if (di[i]!=0) break;            if (i==strlen(big)) cout<<"0";            else                for (j=i; j<strlen(big); j++)                    cout<<di[j];            cout<<endl;        }    }    return 0;}