贪心 BestCoder Round #39 1001 Delete

 

题目传送门

 1 /*  2  贪心水题：找出出现次数>1的次数和res，如果要减去的比res小，那么总的不同的数字tot不会少；  3  否则再在tot里减去多余的即为答案  4  用set容器也可以做，思路一样  5 */  6 #include <cstdio>  7 #include <iostream>  8 #include <cstring>  9 #include <string> 10 #include <algorithm> 11 using namespace std; 12 13 const int MAXN = 1e4 + 10; 14 const int INF = 0x3f3f3f3f; 15 int cnt[110]; 16 17 int main(void) //BestCoder Round #39 1001 Delete 18 { 19 //freopen ("1001.in", "r", stdin); 20 21 int n; 22 while (scanf ("%d", &n) == 1) 23  { 24 int k; 25 memset (cnt, 0, sizeof (cnt)); 26 27 int tot = 0, res = 0, x; 28 for (int i=1; i<=n; ++i) 29  { 30 scanf ("%d", &x); 31 if (cnt[x] == 0) tot++; 32 else if (cnt[x] >= 1) res++; 33 cnt[x]++; 34  } 35 36 scanf ("%d", &k); 37 if (res >= k) printf ("%d\n", tot); 38 else printf ("%d\n", tot - (k - res)); 39  } 40 41 return 0; 42 }

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <cstring>

 4 #include <string>

 5 #include <algorithm>

 6 #include <set>

 7 using namespace std;  8 

 9 int main(void)        //BestCoder Round #39 1001 Delete

10 { 11     //freopen ("1001.in", "r", stdin);

12 

13     set<int> S; 14     int n, k; 15 

16     while (cin >> n) 17  { 18  S.clear (); 19         int x; 20         for (int i=1; i<=n; ++i) 21  { 22             cin >> x; S.insert (x); 23  } 24 

25         cin >> k; 26         cout << ((n-S.size () <= k) ? n - k : S.size ()) << endl; 27  } 28 

29     return 0; 30 }

使用set容器