用时最短，内存最优

状态转移方程： AC代码：

#include<bits/stdc++.h> using namespace std; int main(){ int n,sum=0; cin >> n; sum+=n/11; n=n%11; sum+=n/5; n=n%5; sum+=n; cout << sum; return 0; }

#include<bits/stdc++.h> using namespace std; int n; int l = 0; int dp; void h(){ if(11 <= dp){ dp -= 11; l += 1; }else if(5 <= dp){ dp -= 5; l += 1; }else if(1 <= dp){ dp -= 1; l += 1; } if(dp == 0){ cout << l; return; }else{ h(); } } int main(){ cin >> n; dp = n; h(); return 0; }

#include<iostream> using namespace std; int main() { int coin[3]={1,5,11}; int s; cin>>s; int dp[s+1]={0}; for(int i=0;i<=s;i++) { dp[i]=10000001; } dp[0]=0; for(int i=0;i<=s;i++) { for(int j=0;j<3;j++) { if(coin[j]<=i) { dp[i]=min(dp[i],dp[i-coin[j]]+1); } } } if(dp[s]==10000001) { cout<<"-1"; } else { cout<<dp[s]; } }