Codeforces Round #434 C

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题意：a e i o u是元音，在一个字符串里不能有3个连续的辅音，但是3个及以上相同的连续辅音可以，否则就要用空格隔开，给你字符串，用最少的空格把字符串变为合法

思路：模拟就是了

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("
")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;

char s[30005];
int is[1005],ans[30005];

int main(){
    cin>>s+1;
    is['a']=1, is['e']=1, is['i']=1, is['o']=1, is['u']=1;
    int n=strlen(s+1);
    for(int i=1; i<=n; ++i){
        if(!is[s[i]]){
            int k=0;
            for(int j=i; j<=n; ++j){
                if(is[s[j]]){
                    i=j;
                    break;
                }
                k++;
                if(k>2){
                    if(s[j] == s[j-1] && s[j-1]==s[j-2]){
                        for(int t=j+1; t<=n; ++t){
                            if(s[t]!=s[t-1]){
                                if(is[s[t]]){
                                    i=t;
                                    break;
                                }
                                ans[t-1]=1;
                                i=t-1;
                                break;
                            }
                        }
                    }
                    else{
                        ans[j-1]=1;
                        i=j-1;
                        break;
                    }
                    break;
                }
            }
        }
    }
    for(int i=1; i<=n; ++i){
        cout<<s[i];
        if(ans[i]) cout<<" ";
    }
    return 0;
}