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  • A. Cards (模拟) (Codeforces Round #586 (Div. 1 + Div. 2))

    When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.

    Input

    The first line contains a single integer nn (1n1051⩽n⩽105) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.

    It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 00 or "one" which corresponds to the digit 11.

    Output

    Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.

    Examples
    input
    Copy
    4
    ezor
    
    output
    Copy
    0 
    
    input
    Copy
    10
    nznooeeoer
    
    output
    Copy
    1 1 0 
    
    Note

    In the first example, the correct initial ordering is "zero".

    In the second example, the correct initial ordering is "oneonezero".

     

     

     

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <string>
    #include <cstring>
    #include <cstdlib>
    #include <map>
    #include <vector>
    #include <set>
    #include <queue>
    #include <stack>
    #include <cmath>
    using namespace std;
    #define pq priority_queue<int>
    #define pql priority_queue<ll>
    #define pqn priority_queue<node>
    #define v vector<int>
    #define vl vector<ll>
    #define lson rt<<1, l, m
    #define rson rt<<1|1, m+1, r
    #define read(x) scanf("%d",&x)
    #define lread(x) scanf("%lld",&x);
    #define pt(x) printf("%d
    ",(x))
    #define yes printf("YES
    ");
    #define no printf("NO
    ");
    #define gcd __gcd
    #define out(x) cout<<x<<endl;
    #define over cout<<endl;
    #define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++)
    #define input(k) for (int i = 1; i <= (int)(k); i++)  {cin>>a[i] ; }
    #define mem(s,t) memset(s,t,sizeof(s))
    #define re return ;
    #define ok return 0;
    #define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    #define mod(x) ((x)%9973)
    #define test  cout<<"       ok             "<<endl;
    #define ls now<<1
     
    int main()
    {
        TLE;
        int n,ans,cnt;
        string str;
        while(cin>>n)
        {
            ans=0; cnt=0;
            cin>>str;
            for(int i=0;i<str.size();i++)
                if(str[i]=='z') ans++;
                else if(str[i]=='n') cnt++;
            if(!cnt)
            {
                cout<<0;
                for(int i=1;i<ans;i++)
                    cout<<" "<<0;
                cout<<endl;
     
            }
            else
            {
                cout<<1;
                for(int i=1;i<cnt;i++)
                    cout<<" "<<1;
                for(int i=0;i<ans;i++)
                    cout<<" "<<0;
                cout<<endl;
            }
        }
        ok;
    }

     

     

    所遇皆星河
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  • 原文地址:https://www.cnblogs.com/Shallow-dream/p/11586957.html
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