'Programming Challenges'에 해당되는 글 2건

  1. 2017.07.05 [Programming Challenges] 문제12 암호깨기 (Crypt Kicker)
  2. 2017.07.05 [Programming Challenges] 문제1 3n+1문제
2017.07.05 01:36


문제12 암호깨기 (Crypt Kicker)


A common but insecure method of encrypting text is to permute the letters of the alphabet. In other words, each letter of the alphabet is consistently replaced in the text by some other letter. To ensure that the encryption is reversible, no two letters are replaced by the same letter.

Your task is to decrypt several encoded lines of text, assuming that each line uses a different set of replacements, and that all words in the decrypted text are from a dictionary of known words.

Input

The input consists of a line containing an integer n, followed by n lowercase words, one per line, in alphabetical order. These nwords compose the dictionary of words which may appear in the decrypted text. Following the dictionary are several lines of input. Each line is encrypted as described above.

There are no more than 1,000 words in the dictionary. No word exceeds 16 letters. The encrypted lines contain only lower case letters and spaces and do not exceed 80 characters in length.

Output

Decrypt each line and print it to standard output. If there are multiple solutions, any one will do. If there is no solution, replace every letter of the alphabet by an asterisk.

Sample Input

6
and
dick
jane
puff
spot
yertle
bjvg xsb hxsn xsb qymm xsb rqat xsb pnetfn
xxxx yyy zzzz www yyyy aaa bbbb ccc dddddd

Sample Output

dick and jane and puff and spot and yertle
**** *** **** *** **** *** **** *** ******


# Source (My Solution)

ddddddddddddddddddd
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Posted by injunech
2017.07.05 01:21


문제1 3n+1문제

출처 : http://www.programming-challenges.com/


Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22:

22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000.

For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.

Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.

Output

For each pair of input integers i and j, output ij in the same order in which they appeared in the input and then the maximum cycle length for integers between and including i and j. These three numbers should be separated by one space, with all three numbers on one line and with one line of output for each line of input.

Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174




# Source (My Solution)

#include 

//#define DBG

void main(void) {
	int a;
	int b;

	int tmp;
	int i;
	int x;
	int loop_count;
	int max_count =0;

	scanf("%d", &a);
	scanf("%d", &b);
	
	// If A is Biger than B,  Swap A and B
	if (a > b) {
		tmp = a;
		a = b;
		b = tmp;
	}
	
	for (i = a; i <= b; i++) {
		// Start, Init
		x = i;
		loop_count = 1;
#ifdef DBG
		printf("[%d] =============== \n", i);
#endif

		// Loop
		while (x != 1) {
			if ((x & 1) == 0) {	// even
#ifdef DBG
				printf("[%d] is EVEN\n", x);
#endif
				x = x / 2;
			}
			else {	// odd
#ifdef DBG
				printf("[%d] is ODD\n", x);
#endif
				x = (x * 3) + 1;
			}
			loop_count++;
		}

		if (max_count < loop_count)
			max_count = loop_count;

	}

	printf("%d %d %d", a,b, max_count);
	
}

 


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Posted by injunech