hdu 5210

Delete

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 329    Accepted Submission(s): 197

Problem Description

WLD likes playing with numbers. One day he is playing with N integers. He wants to delete K integers from them. He likes diversity, so he wants to keep the kinds of different integers as many as possible after the deletion. But he is busy pushing, can you help him?

 

Input

There are Multiple Cases. (At MOST 100)

For each case:

The first line contains one integer N(0<N≤100).

The second line contains N integers a1,a2,...,aN(1≤ai≤N), denoting the integers WLD plays with.

The third line contains one integer K(0≤K<N).

 

Output

For each case:

Print one integer. It denotes the maximum of different numbers remain after the deletion.

 

Sample Input

4 1 3 1 2 1

 

Sample Output

3

Hint

if WLD deletes a 3, the numbers remain is [1,1,2],he'll get 2 different numbers. if WLD deletes a 2, the numbers remain is [1,1,3],he'll get 2 different numbers. if WLD deletes a 1, the numbers remain is [1,2,3],he'll get 3 different numbers.

 

Source

BestCoder Round #39 ($)

 

 

以上是題目，就是給你n個數要你刪掉k個，讓剩下的數不相同最多。那就是找到能刪掉而不影響最大的答案的有幾個，就是找出有多少個是前面已經有了的，而如果k比能刪的重複數的數目的總和還大，就從n-sum之後再刪，就是刪不同的數，這時候不同數的數目就變了，輸出答案即可。不知爲何竟然錯了，難道是因爲沒有memset？可是我在後來直接定義數組的呀。。不過現在ac了。這道題我在想，要是數據特別多特別大怎麼算呢，當時就想到map，但是map我不會用。。等會兒我去百度百度吧。

#include<iostream>

#include<cstdio>
#include<cstring>
using namespace std;
#define maxx 105 
int a[maxx];
int main()
{
int n;


while(cin>>n)//忽然在想要是n的輸入是“邪惡的” ，那該怎麼判斷規避，還是現在這樣是不用的呢 
{

memset(a,0,sizeof(a));
int sum=0,k,ans=0,x;//sum代表有幾個不一樣的數 
//ans代表最終輸出，k代表輸入的k值，x用來存每次輸入的數。n代表輸入有幾個數
int i=n;
while(i--)
{
//scanf("%d",&x);
cin>>x;
if(!a[x])
{
sum+=1;
}
a[x]+=1;
}
cin>>k;
if(n-sum>=k)
printf("%d\n",sum);//其實之前一直習慣於cin、cout。不過printf什麼的更快，想習慣改成這樣。 
else
{
//printf("%d\n",sum1+sum2-k);
cout<<n-k<<endl;
}
} 

return 0;

} 

//=============接下來是原來的代碼，試了試發現問題在於memset。。但是我明明是輸入n之後才定義的數組a啊怎麼就不行呢。==========================

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
int a[105],k,sum=0,x;
memset(a,0,sizeof(a));
for(int i=0;i<n;i++)
{
cin>>x;
if(a[x]!=1)//就算只有一行代碼，也要有花括號，要寫得規範，便於閱讀。 
{
a[x]=1;
}
else 
{
sum+=1;
}
}
cin>>k;
if(sum>=k)
{
printf("%d\n",n-sum);
}
else 
{
printf("%d\n",n-k);
}

}
return 0;
}