1 条题解

  • 0
    @ 2025-2-14 21:20:42

    C :

    #include <stdio.h>
    #include <string.h>
    
    #define MAX 20001
    
    int main(void)
    {
       int m,n,tv,v,i,j,k ;
       int a[MAX] ={0} ;
       a[0] = 1;
       scanf("%d",&v) ;
       scanf("%d",&n) ;
     
       
       for(i=1 ; i<= n ; i++)
       {
           scanf("%d",&tv);
           for(j=v ; j>=tv ; j--)
           if(!a[j])
           a[j] = a[j-tv] ;
        }
       m = v ;
       while ( a[m] == 0)
       m -- ;
       printf("%d ",v-m) ;
      
       return 0 ;
    }
    

    C++ :

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    using namespace std;
    int f[50000],w[4000],i,j,n,m;;
    int main ()
    {
    	cin>>m;
    	cin>>n;
    	for (i=1; i<=n; i++)
    	  cin>>w[i];
    	for (i=1; i<=n; i++)
    	  for (j=m; j>=w[i]; j--)
    	f[j]=max(f[j-w[i]]+w[i],f[j]);
    	cout<<m-f[m];
    }
    
    

    Pascal :

    program bb;
    var  w,v:array[0..20005] of longint;
         f:array[0..20005] of longint;
         vv,n,i,j:longint;
    begin
      read(vv,n);
      for i:=1 to n do read(v[i]);
    
      for i:=1 to n do
        for j:=vv downto v[i] do
          if f[j-v[i]]+v[i]>f[j] then f[j]:=f[j-v[i]]+v[i];
      writeln(vv-f[vv]);
    end.
    

    Java :

    
    import java.util.Scanner;
    
    /**
     * Created by ResiakFX on 2017/2/26 0026.
     */
    public class Main {
        public static void main(String[] args){
            Scanner in=new Scanner(System.in);
            while(in.hasNext()){
                int V=in.nextInt();
                int N=in.nextInt();
                int[] a=new int[N];
                for(int i=0;i<N;i++)a[i]=in.nextInt();
                int[][] d=new int[N][V+1];
    
                for(int i=0;i<N;i++){
                    for(int j=1;j<=V;j++){
                        if(i==0)d[0][j]=a[0]>j?0:a[0];
                        else{
                            d[i][j]=d[i-1][j];
                            if(j>=a[i] && d[i-1][j-a[i]]+a[i]>d[i][j]) d[i][j]=d[i-1][j-a[i]]+a[i];
                        }
                    }
                }
                System.out.println(V-d[N-1][V]);
            }
        }
    }
    
    
    • 1

    信息

    ID
    847
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者