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jeff69

Untitled

Mar 26th, 2016
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  1.  
  2. using namespace std;
  3. int A[20] , K ;
  4. int dp[1000];
  5. int F(int n){
  6. if(n==0) return 0;
  7. if(dp[n]!=-1) return dp[n];
  8. dp[n] = (1<<20);
  9. for(int j=0;j<K;j++)
  10. if(A[j] <= n)
  11. dp[n] = min(dp[n] , F(n-A[j])+1);
  12. return dp[n];
  13. }
  14. int main(){
  15. fill(dp , dp+1000 , -1);
  16. int n;
  17. cin>>n;
  18. cout<<F(n)<<endl;
  19. //
  20. }
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