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  1. //BÀI 1 : BẦU CỬ (CODE)
  2.  
  3.  
  4. #include <bits/stdc++.h>
  5.  
  6. using namespace std;
  7.  
  8. int n,k,T;
  9. int a[100000 + 5];
  10. void doc()
  11. {
  12.     cin>>n>>k>>T;
  13.     for(int i = 1; i <= n; i ++ )cin>>a[i];
  14. }
  15.  
  16. void sub1()
  17. {
  18.     int sl1 = 0,slmx = 0,dem = 0;
  19.     // sl1 : la so luong nguoi vote cho 1
  20.     // slmx la so luong nguoi cho nguoi cao co luong vote cao nhat
  21.  
  22.     for(int i = 1; i <= n; i ++)
  23.     {
  24.         if(a[i] == 1)sl1 ++;
  25.     }
  26.  
  27.     for(int i = 2; i <= k ; i++)
  28.     {
  29.         int dem = 0;
  30.         for(int j = 1; j <= n; j++)
  31.         {
  32.             if(a[j] == i)dem++;
  33.         }
  34.         slmx = max(slmx,dem);
  35.     }
  36.  
  37.     if(sl1 >= slmx)cout<<"NO NEED";
  38.     else if(sl1 < slmx && slmx - sl1 <= T)
  39.     {
  40.         cout<<"JUST NEED "<<slmx - sl1;
  41.     }
  42.     else cout<<"IMPOSSIBLE";
  43.  
  44. }
  45.  
  46. int sl[100000 + 5];
  47. void sub2()
  48. {
  49.     for(int i = 1; i <= n;i ++ )sl[a[i]] ++;
  50.  
  51.     int slmx = 0;
  52.  
  53.     for(int i = 2; i <= k;i ++ )slmx = max(slmx,sl[i]);
  54.  
  55.     if(sl[1] >= slmx)cout<<"NO NEED";
  56.     else if(sl[1] < slmx && slmx - sl[1] <= T)
  57.     {
  58.         cout<<"JUST NEED "<<slmx - sl[1];
  59.     }
  60.     else cout<<"IMPOSSIBLE";
  61. }
  62.  
  63. int main() {
  64.  
  65.     doc();
  66.     if(n <= 3e3 && k <= 3e3)sub1();
  67.     else sub2();
  68.  
  69.     return 0;
  70. }
  71.  
Success #stdin #stdout 0.01s 5328KB
comments (?)
stdin
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4 2 3
1 2 2 2
stdout
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JUST NEED 2
https://ideone.com/pLeo8S
language:
C++14 (gcc 8.3)
created:
7 hours ago
visibility:
public

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