Greedy bestcoder Round #39 1001 Delete

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1 /*2 Greedy water problem: Find out the number of occurrences >1 and res, if you want to subtract smaller than res, then the total number of different tot will not be less;3 or subtract the excess in tot as the answer.4 you can do it with a set container, like the idea .5 */6#include <cstdio>7#include <iostream>8#include <cstring>9#include <string>Ten#include <algorithm> One using namespacestd; A  - Const intMAXN = 1e4 +Ten; - Const intINF =0x3f3f3f3f; the intcnt[ the]; -  - intMainvoid)//bestcoder Round #39 1001 Delete - { +     //freopen ("1001.in", "R", stdin); -  +     intN; A      while(SCANF ("%d", &n) = =1) at     { -         intK; -memset (CNT,0,sizeof(CNT)); -  -         inttot =0, res =0, X; -          for(intI=1; i<=n; ++i) in         { -scanf ("%d", &x); to             if(Cnt[x] = =0) tot++; +             Else if(Cnt[x] >=1) res++; -cnt[x]++; the         } *  $scanf ("%d", &k);Panax Notoginseng         if(res >= k) printf ("%d\n", tot); -         Elseprintf ("%d\n", Tot-(K-res)); the     } +  A     return 0; the}

1#include <cstdio>2#include <iostream>3#include <cstring>4#include <string>5#include <algorithm>6#include <Set>7 using namespacestd;8 9 intMainvoid)//bestcoder Round #39 1001 DeleteTen { One     //freopen ("1001.in", "R", stdin); A  -     Set<int>S; -     intN, K; the  -      while(Cin >>N) -     { - s.clear (); +         intx; -          for(intI=1; i<=n; ++i) +         { ACIN >>x; S.insert (x); at         } -  -CIN >>K; -cout << ((N-s.size () <= k)? N-k: S.size ()) <<Endl; -     } -  in     return 0; -}

using the Set container

Greedy bestcoder Round #39 1001 Delete