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  • As Simple as One and Two

    You are given a non-empty string s=s1s2sns=s1s2…sn, which consists only of lowercase Latin letters. Polycarp does not like a string if it contains at least one string "one" or at least one string "two" (or both at the same time) as a substring. In other words, Polycarp does not like the string ss if there is an integer jj (1jn21≤j≤n−2), that sjsj+1sj+2=sjsj+1sj+2="one" or sjsj+1sj+2=sjsj+1sj+2="two".

    For example:

    • Polycarp does not like strings "oneee", "ontwow", "twone" and "oneonetwo" (they all have at least one substring "one" or "two"),
    • Polycarp likes strings "oonnee", "twwwo" and "twnoe" (they have no substrings "one" and "two").

    Polycarp wants to select a certain set of indices (positions) and remove all letters on these positions. All removals are made at the same time.

    For example, if the string looks like s=s="onetwone", then if Polycarp selects two indices 33 and 66, then "onetwone" will be selected and the result is "ontwne".

    What is the minimum number of indices (positions) that Polycarp needs to select to make the string liked? What should these positions be?

    Input

    The first line of the input contains an integer tt (1t1041≤t≤104) — the number of test cases in the input. Next, the test cases are given.

    Each test case consists of one non-empty string ss. Its length does not exceed 1.51051.5⋅105. The string ss consists only of lowercase Latin letters.

    It is guaranteed that the sum of lengths of all lines for all input data in the test does not exceed 1.51061.5⋅106.

    Output

    Print an answer for each test case in the input in order of their appearance.

    The first line of each answer should contain rr (0r|s|0≤r≤|s|) — the required minimum number of positions to be removed, where |s||s| is the length of the given line. The second line of each answer should contain rr different integers — the indices themselves for removal in any order. Indices are numbered from left to right from 11 to the length of the string. If r=0r=0, then the second line can be skipped (or you can print empty). If there are several answers, print any of them.

    Examples
    input
    Copy
    4
    onetwone
    testme
    oneoneone
    twotwo
    
    output
    Copy
    2
    6 3
    0
    
    3
    4 1 7 
    2
    1 4
    
    input
    Copy
    10
    onetwonetwooneooonetwooo
    two
    one
    twooooo
    ttttwo
    ttwwoo
    ooone
    onnne
    oneeeee
    oneeeeeeetwooooo
    
    output
    Copy
    6
    18 11 12 1 6 21 
    1
    1 
    1
    3 
    1
    2 
    1
    6 
    0
    
    1
    4 
    0
    
    1
    1 
    2
    1 11 
    
    Note

    In the first example, answers are:

    • "onetwone",
    • "testme" — Polycarp likes it, there is nothing to remove,
    • "oneoneone",
    • "twotwo".

    In the second example, answers are:

    • "onetwonetwooneooonetwooo",
    • "two",
    • "one",
    • "twooooo",
    • "ttttwo",
    • "ttwwoo" — Polycarp likes it, there is nothing to remove,
    • "ooone",
    • "onnne" — Polycarp likes it, there is nothing to remove,
    • "oneeeee",
    • "oneeeeeeetwooooo".

    感觉挺迷惑的一道题。。。

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <unordered_set>
    #include <unordered_map>
    #include <xfunctional>
    #define ll long long
    #define mod 998244353
    using namespace std;
    int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0} };
    const int maxn = 1e5 + 5;
    const long long inf = 0x7f7f7f7f7f7f7f7f;
    
    int main()
    {
        int t;
        cin >> t;
        while (t--)
        {
            string s;
            cin >> s;
            vector<int> v;
            for (int i = 0; i < s.size(); i++)
            {
                if (s[i] == 't')
                {
                    i++;
                    if (i < s.size() && s[i] == 'w')
                    {
                        i++;
                        if (i < s.size() && s[i] == 'o')
                        {
                            i++;
                            if (i < s.size() && s[i] != 'o')
                                v.push_back(i - 1);
                            else
                            {
                                v.push_back(i - 2);
                            }
                            i--;
                            continue;
                        }
                        i--;
                        continue;
                    }
                    i--;
                    continue;
                }
                if (s[i] == 'o')
                {
                    i++;
                    if (i < s.size() && s[i] == 'n')
                    {
                        i++;
                        if (i < s.size() && s[i] == 'e')
                        {                        
                            if (i-3>=0 && s[i - 3] == 'o')
                                v.push_back(i - 1);
                            else
                                v.push_back(i - 2);
                        }
                        i--;
                    }
                    i--;
                }
            }
            cout << v.size() << endl;
            if (v.size() == 0)
                cout << endl;
            for (int i = 0; i < v.size(); i++)
            {
                cout << v[i] + 1 << " ";
                if (i == v.size() - 1)
                    cout << endl;
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/12348121.html
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