2016huasacm暑假集训训练五 H

题目链接：https://vjudge.net/contest/126708#problem/H

题意：A有一大堆的硬币，他觉得太重了，想花掉硬币去坐的士；的士司机可以不找零，但是的士司机也不会多收零钱。怎么样才能使 A 花的零钱最多。

多重背包模板题

AC代码：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main{
  public static void main(String args[])
{
   int N, M;
   Scanner sc=new Scanner(System.in);
   while(sc.hasNext())
   {
      N=sc.nextInt(); 
      M=sc.nextInt(); 
      if(N==0&&M==0) break;
      int a[]=new int[105]; 
      int c[]=new int [105]; 
      int u[]=new int[100001];
      for (int i=0; i< N; i++) a[i]=sc.nextInt();
      for (int i=0; i< N; i++) c[i]=sc.nextInt();
      int sun=0;
      boolean s[]=new boolean[100001];
      s[0]=true;

      for (int i=0; i< N; i++)
      {
         Arrays.fill(u, 0);
          for (int j=a[i]; j<=M; j++)
          {
             if (!s[j] && s[j-a[i]] && u[j-a[i]]< c[i])
              {
                   s[j]=true;
                   u[j]=u[j-a[i]]+1;
                   sun++;
              }
          }
      }
      System.out.println(sun);
   }
    
  } 
}
class Scanner {
    BufferedReader br;
    StringTokenizer st;

    Scanner(InputStream in) {
        br = new BufferedReader(new InputStreamReader(in));
        eat("");
    }

    private void eat(String string) {
        st = new StringTokenizer(string);
    }

    String nextLine() {
        try {
            return br.readLine();
        } catch (IOException e) {
            return null;
        }
    }

    boolean hasNext() {
        while (!st.hasMoreTokens()) {
            String s = nextLine();
            if (s == null)
                return false;
            eat(s);
        }
        return true;
    }

    String next() {
        hasNext();
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }
       
    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }
}