题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1889
1889. Airport Announcements
Time limit: 1.0 second
Memory limit: 64 MB
Igor was bored of waiting in an airport lounge. Oceanic Airlines, a company he didn‘t like so much, delayed the departure of his flight, so he was late for the connection flight to Petrozavodsk, where
a programming camp was to be held. Now he had to wait for long 300 minutes at the airport. Soon he heard a public announcement. Maybe, his flight had been canceled or, maybe, there were discounts on burgers at a nearby bar—Igor couldn‘t tell for sure. It seemed
that the announcement had been repeated in several languages, but, strangely, there hadn‘t been Russian among them.
Igor recognized the language of some of the phrases he had heard. He assumed that the number of phrases in the announcement had been the same regardless of the language and that the announcement had
been made at most once in each language. Help Igor to find the number of languages in which the announcement was made.
Input
The first line contains the total number n of phrases Igor heard (2 ≤ n ≤ 1 000). In the ith of the following n lines you are given the language of the ith
phrase or the word “unknown” if Igor couldn‘t recognize the language. It is guaranteed that Igor could recognize the language of at least one of the phrases. The name of a language is a string of a length from four to twenty symbols consisting of lowercase
English letters.
Output
Output the number of languages in which the announcement was made. If there are several answers, list them in ascending order. If there is no solution, output the string “Igor is wrong.”
Samples
| input | output |
|---|---|
6 english unknown unknown unknown german unknown |
2 3 6 |
4 english french unknown english |
Igor is wrong. |
3 zulu zulu zulu |
1 |
PS:
题意比较难懂，读了好久， 重复播报的时候是 报的112233这样的，不是123123这样！
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n;
string s[1017];
int judge(int t)
{
for(int i = 0; i < n; i+=t)
{
int tt = -1;
for(int j = i; j < i+t; j++)
{
if(s[j] != "unknown")
{
tt = j;
break;
}
}
if(tt == -1)
continue;
for(int j = i; j < i+t; j++)
{
if(s[j] == "unknown")
continue;
if(s[tt] != s[j])
{
return 0;
}
}
for(int j = i+t; j < n; j++)//只出现一次
{
if(s[tt] == s[j])
{
return 0;
}
}
}
return 1;
}
int main()
{
int ans[1017];
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
cin >> s[i];
}
int k = 0;
for(int i = 1; i <= n; i++)
{
if(n%i==0)//整除才有可能
{
int flag = judge(i);//重复的次数
if(flag)
{
ans[k++] = n/i;
}
}
}
sort(ans,ans+k);
if(k == 0)
{
printf("Igor is wrong.\n");
return 0;
}
printf("%d",ans[0]);
for(int i = 1; i < k; i++)
{
printf(" %d",ans[i]);
}
printf("\n");
return 0;
}