Google Code Jam 2009 Qualification Round Problem C. Welcome to Code Jam
本题的 Large dataset 本人尚未解决。
https://code.google.com/codejam/contest/90101/dashboard#s=p2
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 |
Output |
3 |
Case #1: 0001 |
Solution (For small dataset only):
string welcome = string("welcome to code jam");
int tc = ;
int fo[], cp[];
string input;
int inlen;
void radd(int begin_chr, int last_pos) {
if (begin_chr > ) return;
int cp = fo[begin_chr];
if (last_pos > cp) cp = last_pos;
while (cp < inlen) {
if (input[cp] == welcome[begin_chr]) {
if(begin_chr == ) {
tc++;
if (tc == ) tc = ;
}
radd(begin_chr + , cp);
}
cp++;
}
return;
}
int solve()
{
int wlen = ;
inlen = (int)input.length();
tc = ;
for (int wi = ; wi < wlen; wi++) {
char chr = welcome[wi];
bool found = false;
for (int ii = wi; ii < inlen; ii++) {
if (chr == input[ii]) {
fo[wi] = cp[wi] = ii;
found = true;
break;
}
}
if (!found)
return ;
}
radd(, );
return tc;
}
int main()
{
freopen("in.in", "r", stdin);
freopen("out.out", "w", stdout);
int T;
scanf("%d\n", &T);
if (!T) {
cerr << "Check input!" << endl;
exit();
}
for (int t = ; t <= T; t++) {
cerr << "solving: #" << t << " / " << T << endl;
getline(cin, input);
auto result = solve();
printf("Case #%d: %04d\n", t, result);
}
fclose(stdin);
fclose(stdout);
return ;
}
Google Code Jam 2009 Qualification Round Problem C. Welcome to Code Jam的更多相关文章
- Google Code Jam 2009 Qualification Round Problem B. Watersheds
https://code.google.com/codejam/contest/90101/dashboard#s=p1 Problem Geologists sometimes divide an ...
- Google Code Jam 2009 Qualification Round Problem A. Alien Language
https://code.google.com/codejam/contest/90101/dashboard#s=p0 Problem After years of study, scientist ...
- Google Code Jam Africa 2010 Qualification Round Problem B. Reverse Words
Google Code Jam Africa 2010 Qualification Round Problem B. Reverse Words https://code.google.com/cod ...
- Google Code Jam Africa 2010 Qualification Round Problem A. Store Credit
Google Code Jam Qualification Round Africa 2010 Problem A. Store Credit https://code.google.com/code ...
- [C++]Infinite House of Pancakes——Google Code Jam 2015 Qualification Round
Problem It’s opening night at the opera, and your friend is the prima donna (the lead female singer) ...
- [C++]Standing Ovation——Google Code Jam 2015 Qualification Round
Problem It’s opening night at the opera, and your friend is the prima donna (the lead female singer) ...
- Google APAC----Africa 2010, Qualification Round(Problem C. T9 Spelling)----Perl 解法
原题地址链接:https://code.google.com/codejam/contest/351101/dashboard#s=p2 问题描述: Problem The Latin alphabe ...
- Google APAC----Africa 2010, Qualification Round(Problem B. Reverse Words)----Perl 解法
原题地址链接:https://code.google.com/codejam/contest/351101/dashboard#s=p1 问题描述: Problem Given a list of s ...
- Google APAC----Africa 2010, Qualification Round(Problem A. Store Credit)----Perl 解法
原题地址链接:https://code.google.com/codejam/contest/351101/dashboard#s=p0 问题描述: Problem You receive a cre ...
随机推荐
- C#模拟POST登录cnblogs并发布文章
用到的工具FireFox的Firebugs插件 打开网络功能进行抓包 数据如下 可以得知POST的数据为: __EVENTTARGET=&__EVENTARGUMENT=&__VIEW ...
- RPM软件包管理的查询功能
以后大家升级rpm包的时候,不要用Uvh了! 我推荐用Fvh 前者会把没有安装过得包也给装上,后者只会更新已经安装的包 总结:未安装的加上小写p,已安装的不需要加p 查询q rpm {- ...
- 【云计算】开源的Docker Registry WebUI
kwk/docker-registry-frontend Code Issues 9 Pull requests 6 Wiki ...
- (int),Int32.Parse() 和 Convert.toInt32() 的区别
在 C# 中,(int),Int32.Parse() 和 Convert.toInt32() 三种方法有何区别? int 关键字表示一种整型,是32位的,它的 .NET Framework 类型为 S ...
- 如何调试lua脚本
首先感谢下ZeroBrane Studio. 这里拿cocos2dx/samples/Lua/HelloLua做例子来说明,其他的都是同样道理. 1.下载调试Lua所需的IDE,地址在这.有经济实力的 ...
- iOS CoreData 的级联删除等操作
关于CoreData 的基本操作在网上有一些中文资料,但是这些资料大多没有涉及CoreData的详细操作,只是简单的演示了最基本用法.像级联删除这种最基本的数据库操作都没有提到.今天在网上看到了一些英 ...
- HDU 5317 RGCDQ (数论素筛)
RGCDQ Time Limit: 3000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u Submit Status ...
- git merge和个git rebase的区别
http://stackoverflow.com/questions/16666089/whats-the-difference-between-git-merge-and-git-rebase/16 ...
- 组合数(codevs 1631)
1631 组合数 时间限制: 1 s 空间限制: 256000 KB 题目等级 : 钻石 Diamond 题解 查看运行结果 题目描述 Description 组合数C(N, K)表示 ...
- linux Xtrabackup安装及使用方法
[root@centos01 ~]# rpm -Uvh http://www.percona.com/downloads/percona-release/percona-release-0.0-1.x ...