題目：
In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if there are two Bobs?).
A group of soldiers is effective if and only if their names are different. For example, a group (John, Bob, Limak) would be effective, while groups (Gary, Bob, Gary) and (Alice, Alice) wouldn't.
You are a spy in the enemy's camp. You noticed n soldiers standing in a row, numbered 1 through n. The general wants to choose a group of k consecutive soldiers. For every k consecutive soldiers, the general wrote down whether they would be an effective group or not.
You managed to steal the general's notes, with n - k + 1 strings s1, s2, ..., sn - k + 1, each either "YES" or "NO".
- The string s1 describes a group of soldiers 1 through k ("YES" if the group is effective, and "NO" otherwise).
- The string s2 describes a group of soldiers 2 through k + 1.
- And so on, till the string sn - k + 1 that describes a group of soldiers n - k + 1 through n.
Your task is to find possible names of n soldiers. Names should match the stolen notes. Each name should be a string that consists of between 1 and 10 English letters, inclusive. The first letter should be uppercase, and all other letters should be lowercase. Names don't have to be existing names — it's allowed to print "Xyzzzdj" or "T" for example.
Find and print any solution. It can be proved that there always exists at least one solution.
The first line of the input contains two integers n and k (2 ≤ k ≤ n ≤ 50) — the number of soldiers and the size of a group respectively.
The second line contains n - k + 1 strings s1, s2, ..., sn - k + 1. The string si is "YES" if the group of soldiers i through i + k - 1 is effective, and "NO" otherwise.
Find any solution satisfying all given conditions. In one line print n space-separated strings, denoting possible names of soldiers in the order. The first letter of each name should be uppercase, while the other letters should be lowercase. Each name should contain English letters only and has length from 1to 10.
If there are multiple valid solutions, print any of them.
8 3 NO NO YES YES YES NO
Adam Bob Bob Cpqepqwer Limak Adam Bob Adam
9 8 YES NO
R Q Ccccccccc Ccocc Ccc So Strong Samples Ccc
3 2 NO NO
Na Na Na
後來在草稿紙上畫了畫，大概是因爲這樣做滿足前面而不影響後面。
策略：
對於i...i+k.(i=1,2,n-k+1).
如果是"YES"，那麼我就在i+k處新增加一個名字。
否則，我就讓i+k處的名字等於i處的名字。
對了，還有在處理名字時可以先用數字代替，比較方便，然後再將數字轉化成名字。
code：
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
#include<iostream>
using namespace std;
const int MAXN=55;
int num[MAXN];
int main(void){
int n,k;scanf("%d%d",&n,&k);
char s[5];scanf("%s",s);
int tot=1,i=1;
if(s[0]=='Y'){
for(i=1;i<=k;++i)
num[i]=tot++;
}else{
for(i=1;i<=k-1;++i)
num[i]=tot++;
num[i++]=1;
}
for(int j=2;j<=n-k+1;++j){
char s[5];scanf("%s",s);
if(s[0]=='Y'){
num[i++]=tot++;
}else{
num[i]=num[i-k+1];
++i;
}
}
/*for(int i=1;i<=n;++i)printf("%d\t",num[i]);*/
for(int i=1;i<=n;++i){
string s;
bool flag=false;
while(num[i]){
if(!flag){s+='A'+num[i]%10;flag=true;}
else s+='a'+num[i]%10;
num[i]/=10;
}
cout<<s<<" ";
}
printf("\n");
}