UVA 1610 Party Games

看似简单的所谓坑题，其实不坑，只是需要细致周密的考虑以及造数据debug的能力，而这正是acmer必备的素质。

显然我还不是合格的acmer...这题卡了我好几天。。。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=1100;
const int INF=1<<29;

string s[maxn],ans;
int n;

int main()
{
    freopen("in.txt","r",stdin);
    while(cin>>n,n){
        REP(i,1,n) cin>>s[i];
        sort(s+1,s+n+1);
        string a=s[n/2],b=s[n/2+1];
        int la=a.size(),lb=b.size();
        ans="";
        if(la==lb){
            int i=0;
            int tag=0;
            while(i<la){
                if(a[i]==b[i]) ans+=a[i];
                else{
                    if(i==la-1) ans+=a[i];
                    else ans+=a[i]+1;
                    break;
                }
                i++;
            }
        }
        else if(la<lb){
            int i=0;
            while(i<la){
                if(a[i]==b[i]) ans+=a[i];
                else{
                    if(i==la-1) ans+=a[i];
                    else ans+=a[i]+1;
                    break;
                }
                i++;
            }
        }
        else{
            int tag=0;
            REP(i,0,lb-1){
                if(a[i]==b[i]) ans+=a[i];
                else{
                    if(i==lb-1&&a[i]+1==b[i]) tag=1,ans+=a[i];
                    else ans+=a[i]+1;
                    break;
                }
            }
            if(tag){
                REP(i,lb,la-1){
                    if(a[i]=='Z') ans+=a[i];
                    else{
                        if(i==la-1) ans+=a[i];
                        else ans+=a[i]+1;
                        break;
                    }
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}
/**
4
FRED
SAM
JOE
MARGARET
2
FRED
FREDDIE
2
JOSEPHINE
JERRY
2
LARHONDA
LARSEN
2
AAAA
AAAB
2
AAAAZZZ
BBBB
2
AABAAAZZA
AAAAAA
2
AAAB
AAAAZAZ
2
AZ
B
2
AB
Z
2
AC
BB
2
A
BA
2
AA
B
2
ZZZAZZZZZ
ZZZZ
*/