codeforces 791C Bear and Different Names

题目链接：http://codeforces.com/contest/791/problem/C

题意：一个部队有n个人，然后按顺序1到k，2到k+1。。。。。n-k+1到n如果有重复名字，他会给你"NO"，没有就给你"YES"。让你输出n个人的名字，使得符合要求。

分析：n最大50，于是我定义了string s[55],然后在里边存了50个不一样的英文名字，基本就是Aa，Ab这种的，然后开始构造，刚开始几个NO就输出几个一样的名字，当出现第一个YES的时候，输出不一样的名字，然后把名字数组下标记下来。如果接下来是NO，那就输入第前k-1个名字，如果是YES，那就继续输入一个不一样的名字。最后如果人数不够n，输出不同名字补齐即可。细节做好就行，慢慢调试QAQ（代码比较丑，勿看。）

AC代码：

#include<bits/stdc++.h>
using namespace std;
string s[100]={"","Aa","Ab","Ac","Ad","Ae","Af","Ag","Ah","Ai","Aj","Ak","Al","Am","An","Ao","Ap","Aq","Ar","As","At","Au","Av","Aw","Ax","Ay","Az",
"Ba","Bb","Bc","Bd","Be","Bf","Bg","Bh","Bi","Bj","Bk","Bl","Bm","Bn","Bo","Bp","Bq","Br","Bs","Bt","Bu","Bv","Bw","Bx","By","Bz"
"Ca","Cb","Cc","Cd","Ce","Cf","Cg","Ch","Ci","Cj","Ak","Al","Am","An","Ao","Ap","Aq","Ar","As","At","Au","Av","Aw","Ax","Ay","Az"
};
int a[55];
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n,k;
    string c[55];
    cin>>n>>k;
    int j=1;
    int number=1;
    int p=0;
    a[1]=1;
    for(int i=1;i<=n-k+1;i++){
        cin>>c[i];
        if(i==1) cout<<s[1];
        if(c[i]=="NO"){
            if(c[i-1]=="YES"&&p==0){
                p=1;
                for(int q=2;q<k;q++){
                    j++;
                    number++;
                    a[number]=j;
                    cout<<" "<<s[j];
                }
                number++;
                cout<<" "<<s[a[number-k+1]];
                a[number]=a[number-k+1];
            }
            else {
                number++;
                if(p==1) {
                    cout<<" "<<s[a[number-k+1]];
                    a[number]=a[number-k+1];
                }
                else {
                    if(number<k) {
                        cout<<" "<<s[a[number-1]];
                        a[number]=a[number-1];
                    }
                    else {
                        cout<<" "<<s[a[number-k+1]];
                        a[number]=a[number-k+1];
                    }
                 }
            }
        }
        else{
            j++;
            cout<<" "<<s[j];
            number++;
            a[number]=j;
        }
    }
    while(number<n){
        j++;
        number++;
        cout<<" "<<s[j];
    }
    cout<<endl;


return 0;
}