2015年7月20日 星期一

[C][瘋狂程設][CPE考題20141223] 14G3.UVA264:Count on Cantor

Count on Cantor

One of the famous proofs of modern mathematics is Georg Cantor's demonstration that the set of rational numbers is enumerable. The proof works by using an explicit enumeration of rational numbers as shown in the diagram below.
displaymath27
In the above diagram, the first term is 1/1, the second term is 1/2, the third term is 2/1, the fourth term is 3/1, the fifth term is 2/2, and so on.

Input and Output

You are to write a program that will read a list of numbers in the range from 1 to tex2html_wrap_inline29 and will print for each number the corresponding term in Cantor's enumeration as given below. No blank line should appear after the last number.
The input list contains a single number per line and will be terminated by end-of-file.

Sample input

3
14
7

Sample output

TERM 3 IS 2/1
TERM 14 IS 2/4
TERM 7 IS 1/4




Code:


#include <stdio.h>

int main(void){
 int input;
 int num,i,level,sum,value,x,y;
 
 while(scanf("%d",&input)!=EOF){
  num=input;
  i=sum=level=value=0;
  
  while(sum<num){
   i++;
   sum+=i;
   level++;
  }
  value=level+1;
  
  //printf("i=%d,level=%d,sum=%d\n",i,level,sum);
  
  if(level%2==0){
   y=1+(sum-num);
   x=value-y;
  }else{
   x=1+(sum-num);
   y=value-x;
  }
  
  printf("TERM %d IS %d/%d\n",input,x,y);
 }

 return 0;
}

1 則留言:

  1. #include
    void printCantor(int n)
    {
    int m = n, i = 0;
    while(++i < n) n -= i;
    printf("%d\t%d/%d\n", m, n, i+1-n);
    }
    int main()
    {
    for(int i=0;++i!=20;)printCantor(i);
    }

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