uva10494 - 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

高精度除以单精度，

求商思路比较简单直接模拟

求余看了别人的思路才知道自己当初想再用高精度乘法减法算多么的s13

可以边转换边计要计算只包含加法、减法和乘法的整数表达式除以正整数n的余数，可以在每步计算之后对n取余，结果不变。

#include <stdio.h>
#include <string.h>
void main()
{char s[1000],lable;
 long num,i,l,temp,a[1000],sum,pos;
 while (scanf("%s %c %ld",&s,&lable,&num)!=EOF)
 {l=strlen(s); sum=0;
  if (lable=='/')
  {
   temp=0;
   for (i=0;i<l;i++)
   {temp=temp*10+s[i]-'0';
    if (temp>=num)
    {++sum;
     a[sum]=temp/num;
  temp=temp%num;
    }
    else
    {++sum; a[sum]=0;}
   }
   pos=1;
   while ((a[pos]==0)&&(pos<sum)) ++pos;
   for (i=pos;i<=sum;i++)
   printf("%ld",a[i]);
   printf("\n");
  }
  else
  {
   for (i=0;i<l;i++)
   {sum=sum*10+s[i]-'0';
    sum=sum%num;
   }
   printf("%ld\n",sum);
  }
 }
}