UVa 202 large number of divisions

Background: 1_wa: Forget that there is a blank line between each answer! 2_wa: I don't see a space on either side of the equals sign!!!!!!!!

Idea: integers and decimals are separated, integers are divided directly by integer division, fractional part: numerator = (numerator% denominator) *10. and storing each molecule in str[0], the fractional part begins to circulate when there are molecules that have already appeared!

Learn:

1. Pigeon principle (drawer principle, Dirichlet principle):

Simple description: If there are n cages, n+1 only pigeons live, then at least one cage has two pigeons.

A generalized description ( described with a Gaussian function ): Divide n elements into M-sets, at least one set of elements greater than or equal to [(n-1)/m]+1.

For the subject, a%b up to b-1 a case, according to the principle of nest pigeons, the number of links is less than equal to 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);  }  

UVa 202 large number of divisions