NewFolder(1)

标签：NewFolder int Sample newfolder Input ldots newfile

Problem Statement

For two strings $A$ and $B$, let $A+B$ denote the concatenation of $A$ and $B$ in this order.

You are given $N$ strings $S_1,\ldots,S_N$. Modify and print them as follows, in the order $i=1, \ldots, N$:

if none of $S_1,\ldots,S_{i-1}$ is equal to $S_i$, print $S_i$;
if $X$ $(X>0)$ of $S_1,\ldots,S_{i-1}$ are equal to $S_i$, print $S_i+$ ( $+X+$ ), treating $X$ as a string.

Constraints

$1 \leq N \leq 2\times 10^5$
$S_i$ is a string of length between $1$ and $10$ (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

$N$
$S_1$
$S_2$
$\vdots$
$S_N$

Output

Print $N$ lines as specified in the Problem Statement.

Sample Input 1

5
newfile
newfile
newfolder
newfile
newfolder

Sample Output 1

newfile
newfile(1)
newfolder
newfile(2)
newfolder(1)

Sample Input 2

11
a
a
a
a
a
a
a
a
a
a
a

Sample Output 2

a
a(1)
a(2)
a(3)
a(4)
a(5)
a(6)
a(7)
a(8)
a(9)
a(10)

拿个map对字符串进行统计即可。

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
string s[N];
char t[15];
int n;
map<string,int>g;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",t);
		s[i]=t;
		printf("%s",t);
		if(g[s[i]])
			printf("(%d)",g[s[i]]);
		putchar('\n');
		g[s[i]]++;
	}
}

标签：NewFolder,int,Sample,newfolder,Input,ldots,newfile
From： https://www.cnblogs.com/mekoszc/p/16656121.html