Google Code Jam Qualification Round 2009 (4)

広告

C問題:Welcome to Code Jam

Problem

So you’ve registered. We sent you a welcoming email, to welcome you to code jam. But it’s possible that you still don’t feel welcomed to code jam. That’s why we decided to name a problem “welcome to code jam.” After solving this problem, we hope that you’ll feel very welcome. Very welcome, that is, to code jam.

If you read the previous paragraph, you’re probably wondering why it’s there. But if you read it very carefully, you might notice that we have written the words “welcome to code jam” several times: 400263727 times in total. After all, it’s easy to look through the paragraph and find a ‘w’; then find an ‘e’ later in the paragraph; then find an ‘l’ after that, and so on. Your task is to write a program that can take any text and print out how many times that text contains the phrase “welcome to code jam”.

To be more precise, given a text string, you are to determine how many times the string “welcome to code jam” appears as a sub-sequence of that string. In other words, find a sequence s of increasing indices into the input string such that the concatenation of input[s[0]], input[s[1]], …, input[s[18]] is the string “welcome to code jam”.

The result of your calculation might be huge, so for convenience we would only like you to find the last 4 digits.

Input

The first line of input gives the number of test cases, N. The next N lines of input contain one test case each. Each test case is a single line of text, containing only lower-case letters and spaces. No line will start with a space, and no line will end with a space.

Output

For each test case, “Case #x: dddd”, where x is the case number, and dddd is the last four digits of the answer. If the answer has fewer than 4 digits, please add zeroes at the front of your answer to make it exactly 4 digits long.

Limits

1 ≤ N ≤ 100
Small dataset

Each line will be no longer than 30 characters.
Large dataset

Each line will be no longer than 500 characters.

Sample

Input
3
elcomew elcome to code jam
wweellccoommee to code qps jam
welcome to codejam

Output
Case #1: 0001
Case #2: 0256
Case #3: 0000

コメント

最初の3が3行分という意味。各々の行から”welcome to code jam”を何パターン作れるか?という問題。最初の入力ではelcomewのwまでは出番がない。あとはスペースの位置とかおかしいけど、そのままの並びになっているから1通り。

次が面倒くさい。wweellccoommeeと重なっているだけだから計算するのは簡単で、wが2つ、eが2つ・・・・codeの部分はcが1つ、oが1つ・・・・スペースが2つ、あとは1つずつということで、2 x 2 x 2 x ….. x 2 = 256通り。なんだけど、綺麗に並んでいなかったら面倒くさいなと思って再帰でごりごり書くことに。

問題を見ていて、正規表現とかオートマトン(何とかを受理する言語のクラスとか聞いた記憶が)っぽいなあと思ったけど、正規表現ライブラリでどうにかなる感じはしなかった。正規表現の実装したことのある人なら楽勝なんだろうなあと思ったが、やったことがないのだから仕方がない。

ばっちぃコード

#include <stdio.h>
#include <stdlib.h>
#include <string>

//#define DEBUG

#ifdef DEBUG
#endif

using namespace std;

const string welcome = "welcome to code jam";

int answer;


void recursive( string s, int index )
{
	if( welcome.length() <= index )
	{
#ifdef DEBUG
		puts("end of recursive");
#endif
		answer++;
		return;
	}

#ifdef DEBUG	
	printf("recursive : "%s", index = %dn", s.c_str(), index );
#endif
	
	while(true)
	{
		char c = welcome[index];
#ifdef DEBUG
		printf("looking for %cn", c);
#endif
		int pos = s.find(c);
		if( pos != string::npos )
		{
			recursive( s.substr( pos + 1 ), index + 1 );
			s[pos] = '*';
		}
		else
		{
			break;
		}		
	}

}

int main()
{
	FILE *in = fopen("input.txt", "rt");
	char line[8192];
	int N;
	
	if(in == NULL)
		return EXIT_FAILURE;
	
	
	fgets(line, sizeof(line), in);
	sscanf(line, "%d", &N );
	
#ifdef DEBUG
	printf("N = %dn", N );
#endif
	
	for( int i = 0; i < N; i++ )
	{
		string characters;
		string formatted = "";
		
		fgets(line, sizeof(line), in);
		for( int i = 0; i < sizeof(line); i++ )
			if( line[i] == 'n' )
				line[i] = '';
		
		
		characters = line;
#ifdef DEBUG
		puts(characters.c_str());
#endif
		
		answer = 0;
		recursive(characters, 0);

		answer %= 1000;
		printf("Case #%d: %04dn", i + 1, answer );
		answer = 0;		
	}	
}

サンプルとsmallはすぐに答えがでて、correctになった。largeをやると・・・・時間がかかりすぎる。これは再帰では無理だったかと思うけど、後の祭り。結局、時間切れでCのlargeは取れませんでした。