UVA 264 - Count on Cantor

最新推荐文章于 2018-02-16 22:39:57 发布

chiiis

最新推荐文章于 2018-02-16 22:39:57 发布

阅读量192

点赞数

分类专栏： UVA

UVA 专栏收录该内容

68 篇文章

订阅专栏

本文探讨了Georg Cantor的著名数学证明之一，即有理数的可数性。通过直观的图示，展示了如何按Cantor的顺序枚举有理数。文章提供了两种解决方法，用于根据输入的数字输出对应的Cantor序列项。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 toand 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

Solution1:

#define RUN
#ifdef RUN

#include<stdio.h>


// 1：第i条斜线有i个元素
// 2：前i条斜线共有k*(k+1)/2个数
// 3：若存在最小正整数k使得S(k)>=n，
// 则n是第k条斜线上的倒数第S(k)-n+1个数，
// 对应的值是S(k)-n+1 / k-S(k)+n
int main() {

#ifndef ONLINE_JUDGE
	freopen("264.in", "r", stdin);
	freopen("264.out", "w", stdout); 
#endif

  long n;
  while(scanf("%ld", &n) == 1) {
    long k = 1, s = 0;
    for(;;) {
      s += k;
	  
	  // 找到了最小正整数k
      if(s >= n) {
		if(k&1){
			printf("TERM %ld IS %ld/%ld\n", n, s-n+1, k-s+n);
		}
		else{
			printf("TERM %ld IS %ld/%ld\n", n, k-s+n, s-n+1);
		}
        break;
      }
      k++;
    }
  }
  return 0;
}

#endif

Solution2:

//#define RUN
#ifdef RUN

#include<stdio.h>
#include<math.h>
int main() {

#ifndef ONLINE_JUDGE
	freopen("264.in", "r", stdin);
	freopen("264.out", "w", stdout); 
#endif

  long n;
  while(scanf("%ld", &n) == 1) {
    long k = (long)floor((sqrt(8.0*n+1)-1)/2-1e-9)+1;
    long s = k*(k+1)/2;

	if(k&1){
		printf("TERM %ld IS %ld/%ld\n", n, s-n+1, k-s+n);
	}
	else{
		printf("TERM %ld IS %ld/%ld\n", n, k-s+n, s-n+1);
	}



  }
  return 0;
}

#endif