Codeforces189 A. Cut Ribbon(DP)

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给你一个数n，和三个数a，b，c；

可以从a，b，c中任意取任意次，问最多能用多少个a,b,c组成n，一定有结果。

直接搜索一发。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#define PI acos(-1.0)
#define maxn (20000 + 50)
#define MOD 1000000009
#define INF 0x3f3f3f3f
#define Lowbit(x) (x & (-x))
#define mem(a,x) memset(a,x,sizeof(a))
#define Read()  freopen("in.txt", "r", stdin);
#define Write() freopen("out.txt", "w", stdout);
#define bitnum(a) __builtin_popcount(a)
using namespace std;
typedef long long int ll;
inline int in()
{
    int res=0;char c;int f=1;
    while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
int dp[4444];
int a,b,c,n;int maxx = 1;
void dfs(int sum,int cnt){<pre name="code" class="cpp">int main(){
    int n,a,b,c;
    cin>>n>>a>>b>>c;
    int f[4444];
    for(int i=1;i<=n;i++){
        //cout<<f[i]<<" ";
        f[i] = -10000;
        if(i>=a && f[i-a] +1 > f[i])
            f[i] = f[i-a] + 1;
        if(i >= b&& f[i-b] +1 > f[i])
            f[i] = f[i-b] + 1;
        if(i >= c&& f[i-c] +1 > f[i])
            f[i] = f[i-c] + 1;
       // cout<<f[i]<<endl;
    }
    printf("%d\n",f[n]);
}
int main(){
    int n,a,b,c;
    int dp[4444];
    cin>>n>>a>>b>>c;
    mem(dp,-1);
    dp[0] = 0;
    for(int i=0;i<=n;i++)
    if (dp[i] != -1){
        dp[i+a] = max(dp[i+a],dp[i] + 1);
        dp[i+b] = max(dp[i+b],dp[i] + 1);
        dp[i+c] = max(dp[i+c],dp[i] + 1);
    }
    cout<<dp[n]<<endl;
}

if(sum == n){ maxx = max(cnt,maxx); } else if(sum > n) return; if(!dp[sum+a] || dp[sum+a] < cnt+1){ dp[sum+a] = cnt+1; dfs(sum+a,cnt+1); } if(!dp[sum+b] || dp[sum+b] < cnt+1){ dp[sum+b] = cnt+1; dfs(sum+b,cnt+1); } if(!dp[sum+c] || dp[sum+c] < cnt+1){ dp[sum+c] = cnt+1; dfs(sum+c,cnt+1); }}
int main(){

    n = in();
    a = in(); b = in();c = in();
    dfs(0,0);
    printf("%d\n",maxx);
    return 0;
}

代码太丑，接着DP两发

int main(){
    int n,a,b,c;
    cin>>n>>a>>b>>c;
    int f[4444];
    for(int i=1;i<=n;i++){
        //cout<<f[i]<<" ";
        f[i] = -10000; //标记<span style="display: none; width: 0px; height: 0px;" id="transmark"></span>
        if(i>=a && f[i-a] +1 > f[i])
            f[i] = f[i-a] + 1;
        if(i >= b&& f[i-b] +1 > f[i])
            f[i] = f[i-b] + 1;
        if(i >= c&& f[i-c] +1 > f[i])
            f[i] = f[i-c] + 1;
       // cout<<f[i]<<endl;
    }
    printf("%d\n",f[n]);
}
int main(){
    int n,a,b,c;
    int dp[4444];
    cin>>n>>a>>b>>c;
    mem(dp,-1);
    dp[0] = 0;
    for(int i=0;i<=n;i++)
    if (dp[i] != -1){
        dp[i+a] = max(dp[i+a],dp[i] + 1);
        dp[i+b] = max(dp[i+b],dp[i] + 1);
        dp[i+c] = max(dp[i+c],dp[i] + 1);
    }
    cout<<dp[n]<<endl;
}

最后暴力一发。

int main()
{
  int N,A,B,C;cin>>N>>A>>B>>C;
  int Ans=0;
  for(int i=0;i*A<=N;i++)
    for(int j=0;i*A+j*B<=N;j++)
      {
        int k=(N-i*A-j*B)/C;
        if(i*A+j*B+k*C==N)
          Ans=max(Ans,i+j+k);
      }
  cout<<Ans<<endl;
  return 0;
}