Description
There are many users on SDNU OJ, but only the one who registered can use it. We think different username represent different user.
One day ZhouXiaohui got a record of "Status Pages", the captain Albert_s wants to know how many users are using SDNU OJ at that time.
He will give you an "Accepted" as a gift if you help him figure out this problem.
Input
The first line of the input contains an integer T, denoting the number of "Status Pages" ZhouXiaohui got.In each "Status Pages", there is a single integers n in one line, denoting all the username on this pages, then following username ., ,
Output
For each testcase, print a single line starting with "Case #i:"(i indicates the case number) and then a single integer, denoting the answer Albert_s wants to know, separated with a single space.
Sample Input
1
3
ZhouXiaohui
ZhouLaohui
ZhouXiaohui
Sample Output
Case #1: 2
Source
Unknown
#include <cstdio>
#include <iostream>
#include <
string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long
int t, n, sum;
string name[
100000+
8];
int main()
{
scanf("%d", &
t);
int miao =
t;
while(t--
)
{
bool flag =
0;
int num =
0;
cin>>
n;
for(
int i =
0; i<n; i++
)
{
flag =
0;
cin>>
name[i];
for(
int j =
0; j<i; j++
)
{
if(name[i] ==
name[j])
{
flag =
1;
break;
}
}
if(flag)
continue;
else num++
;
}
printf("Case #%d: %d\n", miao-
t, num);
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/11218141.html
相关资源：JAVA上百实例源码以及开源项目
