Party Games(uva 1610) 卡细节

 You’ve been invited to a party. The hostwants to divide the guests into 2 teams for party games, with exactly the samenumber of guests on each
team. She wants to be able to tell which guest is onwhich team as she greets them when they arrive. She’d like to do so as easilyas possible, without having to take the time to look up each guest’s name on alist. Being a good computer scientist, you have an
idea: give her a singlestring, and all she has to do is compare the guest’s name alphabetically tothat string. To make this even easier, you would like the string to be as shortas possible. Given the unique names of n party guests (n is even), find theshortest
possible string S such that exactly half the names are less than orequal to S, and exactly half are greater than S. If there are multiple stringsof the same shortest possible length, choose the alphabetically smallest stringfrom among them.

 

Input 

 There may be multiple test cases inthe input. Each test case will begin with an even integer n (2 ≤ n ≤ 1, 000) onits own line. On the next n
lines will be names, one per line. Each name willbe a single word consisting only of capital letters and will be no longer than30 letters. The input will end with a ‘0’ on its own line.

 

Output 

 For each case, print a single linecontaining the shortest possible string (with ties broken in favor of thealphabetically smallest) that your
host could use to separate her guests. Thestrings should be printed in all capital letters.

 

Sample Input

4

FRED

SAM

JOE

MARGARET

2

FRED

FREDDIE

2

JOSEPHINE

JERRY

2

LARHONDA

LARSEN 

0

 

Sample Output

K

FRED

JF

LARI

 

题意

 给定偶数n个字符串，要求找出一个最短的字符串使得有一半的字符串字典序小于等于该串，另一半大于该串，保证有解。

 

思路

 思路很简单，就是给字符串排序考虑最中间的两个串就行。别的不多说，给几组坑爹的数据就会做了。卡了2小时想哭。

 

Sample input

2

ABCDEFGH

AC

Sample output

ABD

 

Sample input

2

XXXXA

XXXXB

Sample output

XXXXA

 

Sample input

2

XXXXAHHHH

XXXXBP

Sample output

XXXXB

 

Sample input

2

XXXXAHHHH

XXXXB

Sample output

XXXXAI

 

Sample input

2

XXXXAZZZZ

XXXXB

Sample output

XXXXAZZZZ

 

Sample input

2

XXXXAZZZZHKLJ

XXXXB

Sample output

XXXXAZZZZI

 

最坑的是这个：

Sample input

2

XXXXAZZZZZA

XXXXB

Sample output

XXXXAZZZZZA

 

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 50;

int n;
vector<string> s;
char ans[maxn], in[maxn];

int main() {
while (scanf("%d", &n) == 1 && n) {

s.clear();
memset(ans, 0, sizeof(ans));

for (int i = 0; i < n; i++) {
scanf("%s", in);
s.push_back(string(in));
}

sort(s.begin(), s.end());

int i;
string s1 = s[n/2 - 1], s2 = s[n/2];
int len1 = s1.length(), len2 = s2.length();

if (len1 <= len2) {
for (i = 0; i < len1; i++) {
if (s1[i] == s2[i]) ans[i] = s1[i];
else break;
}
if (i != len1) {
if (i + 1 == len1) ans[i] = s1[i];
else ans[i] = s1[i] + 1;
}
}
else {
for (i = 0; i < len2; i++) {
if (s1[i] == s2[i]) ans[i] = s1[i];
else break;
}
if (i + 1 == len2) {
if (s1[i] + 1 < s2[i]) ans[i] = s1[i] + 1;
else {
ans[i] = s1[i];
for (i++; i < len1; i++) {
if (s1[i] == 'Z') ans[i] = s1[i];
else break;
}
if (i < len1 - 1) ans[i] = s1[i] + 1;
else if (i == len1 - 1) ans[i] = s1[i];
}
}
else { ans[i] = s1[i] + 1; }
}
printf("%s\n", ans);
}
return 0;
}