The method of dividing a large number by a maximum of 202 by using a uva202 division by a large number.

The method of dividing a large number by a maximum of 202 by using a uva202 division by a large number.

Background: vodwa: forget that there is a blank line between each answer! 2_WA: no space is displayed on both sides of the equal sign !!!!!!!!

Idea: Separate integers and decimals. integer division is used directly. fractional part: Numerator = (numerator % denominator) * 10. in addition, each molecule is stored in str [0]. When an existing molecule appears, the fractional part starts to loop!

Learning:

1. Nest pigeon principle (Drawer principle, Dirichlet principle ):

Simple Description: if there are n cages where n + 1 pigeons reside, at least one cage contains two pigeons.

General description (Describe using Gaussian Functions): Divides n elements into m sets. At least one set contains more than or equal to [(n-1)/m] + 1.

For this question, a % B has a maximum of B-1, according to the nest pigeon principle, the number of cycles is less than or equal to the B-1.

#include<stdio.h>int str[2][10000];  int main(void){  int a,b,intger;  while(scanf("%d %d",&a,&b)!=EOF){  int count=0,start,aa=a;  intger=a/b;  a=a%b;  while(1){  a*=10;  for(int i=0;i < count;i++){  if(a==str[0][i]){  start=i;            goto l1;  }  }  str[0][count]=a;  str[1][count++]=a/b;  a=a%b;  }l1:   printf("%d/%d = %d.",aa,b,intger);      for(int i=0;i < start;i++){      printf("%d",str[1][i]);      }             if(count<=50){      printf("(");      for(int i=start;i < count;i++){      printf("%d",str[1][i]);      }      printf(")\n");      }else{      printf("(");      for(int i=start;i < start+50;i++){      printf("%d",str[1][i]);      }      printf("...)\n");      }      printf("   %d = number of digits in repeating cycle\n\n",count-start);  }  return 0;}