1563. Bayan
Time limit: 1.0 second
Memory limit: 64 MB
As everybody knows, there are a lot of stores in skyscrapers, it‘s the favourite place of glamorous girls. Blonde Cindy loves only one thing — the shopping. Today is one of the best days, she‘s going shopping to the new skyscraper
“Prisma”. At first she decided to visit all the stores. But actually the “Prisma” is so large that you can find not just one store of each brand. Every time when Cindy found the brand, she has visited before, she told “Bayan”, and went on shopping.
Cindy saw all the stores in the “Prisma”. So how many times did she tell “Bayan”?
Input
First line contains single integer N representing the number of stores in the “Prisma” (1 ≤ N ≤ 1000). In each of nextN lines the brand of store is written. The brands are the strings of Latin letters
and blanks. The length of the string is from 1 to 30. There are no brands, that differ only in register.
Output
Print the number of stores, Cindy didn‘t visit.
Sample
| input | output |
|---|---|
12 ESPRIT Nice Connection Camelot Adilisik Lady and Gentleman City MEXX Camelot Sultanna Frantsuzova Camaieu MEXX Axara Camelot |
3 |
Problem Author: Vladimir Yakovlev
Problem Source: The XIIth USU Programing Championship, October 6, 2007
解析：当一个字符串在之前已经出现过时，说一次“Bayan”。用getline（）或者gets（）都可以读，要注意空格和换行。
PS：之前一直wrong，原来是题意理解错了。。。
AC代码：
#include <cstdio>
#include <map>
#include <string>
#include <iostream>
using namespace std;
map<string, int> m;
int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif //sxk
int n;
string s;
while(scanf("%d", &n)==1){
getchar();
int ans = 0;
m.clear();
for(int i=0; i<n; i++){
getline(cin, s);
if(m.count(s)) ans ++; //出现过
m[s] ++;
}
printf("%d\n", ans);
}
return 0;
}