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/codejam/contest/351101/dashboard#s=p1
Problem
Given a list of space separated words, reverse the order of the words. Each line of text contains L
letters and W
words. A line will only consist of letters and space characters. There will be exactly one space character between each pair of consecutive words.
Input
The first line of input gives the number of cases, N.
N test cases follow. For each test case there will a line of letters and space characters indicating a list of space separated words. Spaces will not appear at the start or end of a line.
Output
For each test case, output one line containing "Case #x: " followed by the list of words in reverse order.
Limits
Small dataset
N = 5
1 ≤ L ≤ 25
Large dataset
N = 100
1 ≤ L ≤ 1000
Sample
Input |
Output |
3 |
Case #1: test a is this |
Solution:
vector<string> solve1(vector<string>words) { reverse(words.begin(), words.end()); return words; } int main() { freopen("in", "r", stdin); //freopen("out", "w", stdout); int t_case_num; scanf("%d\n", &t_case_num); if (!t_case_num) { cerr << "Check input!" << endl; exit(0); } // Read input set for (int case_n = 1; case_n <= t_case_num; case_n++) { string line; getline(cin, line); stringstream ss(line); vector<string>words; string w; while (ss >> w) { words.push_back(w); } auto result = solve1(words); printf("Case #%d: ", case_n); for (int i = 0; i < result.size(); i++) { cout << result.at(i) << " "; } printf("\n"); } fclose(stdin); fclose(stdout); return 0; }