标签：net cte 循环 length sdn for循环 ted minimum case
time limit per test3 seconds
memory limit per test256 megabytes
input: standard input
output: standard output
You are given a non-empty string s=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 s if there is an integer j(1≤j≤n−2), that sjsj+1sj+2=“one” or 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=“onetwone”, then if Polycarp selects two indices 3 and 6, 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 t(1≤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 s. Its length does not exceed 1.5⋅105. The string s 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 1.5⋅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 r(0≤r≤|s|) — the required minimum number of positions to be removed, where |s| is the length of the given line. The second line of each answer should contain r different integers — the indices themselves for removal in any order. Indices are numbered from left to right from 1 to the length of the string. If r=0, then the second line can be skipped (or you can print empty). If there are several answers, print any of them.
Examples
Input
4
onetwone
testme
oneoneone
twotwo
Output
2
6 3
0
3
4 1 7
2
1 4
Input
10
onetwonetwooneooonetwooo
two
one
twooooo
ttttwo
ttwwoo
ooone
onnne
oneeeee
oneeeeeeetwooooo
Output
6
18 11 12 1 6 21
1
1
1
3
1
2
1
6
0
1
4
0
1
1
2
1 11
题目大意：
给出一个字符串，删除部分字符(也可以不删)，使字符没有连续的"one"或者"two"。
最近cf某场的div2的c题，一开始的策略出了点问题，导致连wa三发，罚时瞬间爆炸。策略很简单，如果是单独的"one"就把n删了，如果是单独的"two"就把w删了，就可以使类似"oooneee"这样的多个头跟尾字符的情况。而如果是"twone"，就把o删了。第一次wa是因为没有做判断，for循环的指针也没有跳跃，使"twone"把o删了之后循环到"one"又把n给删了。之后加了个标识数组，蜜汁wa，最后直接把指针做了下跳跃处理就ac了。(脑子不清醒别随便打题.jpg)
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int t;
string s;
int a[100005];
int times,num;
int main(){
cin>>t;
while(t--){
cin>>s;
times=0; num=0;
for(int i=0;i<s.size();i++){
if(s[i]==‘o‘ && s[i+1]==‘n‘&& s[i+2]==‘e‘){
times++;
a[num++]=i+2;
i=i+2;
}else if(s[i]==‘t‘ && s[i+1]==‘w‘ && s[i+2]==‘o‘){
if(s[i+3]==‘n‘ && s[i+4]==‘e‘){
times++;
a[num++]=i+3;
i=i+4;
}else{
times++;
a[num++]=i+2;
i=i+2;
}
}
}
cout<<times<<endl;
for(int i=0;i<num;i++){
cout<<a[i];
if(i!=num-1){
cout<<" ";
}
}
cout<<endl;
}
return 0;
}
————————————————
CSDN链接：https://blog.csdn.net/weixin_43880627/article/details/103621781 As Simple as One and Two
标签：net cte 循环 length sdn for循环 ted minimum case
原文地址：https://www.cnblogs.com/jjmmboom/p/12075373.html
