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  • 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
    this is a test
    foobar
    all your base
    Case #1: test a is this
    Case #2: foobar
    Case #3: base your all

    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
    ", &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("
    ");
            
        }
        
        fclose(stdin);
        fclose(stdout);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/fatlyz/p/3678851.html
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