zoukankan      html  css  js  c++  java
  • Hdu 2844 Coins

    Coins

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 14884    Accepted Submission(s): 5903

    Problem Description

    Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.

    You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

    Input

    The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.

    Output

    For each test case output the answer on a single line.

    Sample Input

    3 10

    1 2 4 2 1 1

    2 5

    1 4 2 1

    0 0

    Sample Output

    8

    4

    #include<string.h>
    #include<stdio.h>
    int n,v;
    int f[100050],val[105],amount[105];
    void zeroOnePack(int cost,int weight)
    {
        for(int i=v;i>=cost;i--)
            if(f[i-cost]+weight>f[i])
                f[i]=f[i-cost]+weight;
    }
    void completePack(int cost,int weight)
    {
        for(int i=cost;i<=v;i++)
        if(f[i-cost]+weight>f[i])
                f[i]=f[i-cost]+weight;    
    } 
    void mutiplePack(int cost,int weight,int amount)
    {
        if(cost*amount>=v)completePack(cost,weight);
        else
        {
            for(int k=1;k<amount;)
            {
                zeroOnePack(k*cost,k*weight);
                amount-=k;
                k=k*2;
            }
            zeroOnePack(amount*cost,amount*weight);
        }
    }
    int main()
    {
    
        while(scanf("%d %d",&n,&v),n,v)
        {
            memset(f,0,sizeof(f));
            memset(val,0,sizeof(val));
            memset(amount,0,sizeof(amount));
            int sum=0;
            for(int i=1;i<=n;i++)
                scanf("%d",&val[i]);
            for(int j=1;j<=n;j++)
                scanf("%d",&amount[j]);
            for(int i=1;i<=n;i++)
                mutiplePack(val[i],val[i],amount[i]);
            for(int i=1;i<=v;i++)
                if(f[i]==i)
                    sum++;
            printf("%d
    ",sum);
        }
        return 0;
    }
    

      

  • 相关阅读:
    NOJ-1581 筷子 (线性DP)
    UVA-242 Stamps and Envelope Size (DP)
    POJ 1860 (SPFA判断正环)
    POJ 3268 最短路水题
    STL----priority_queue
    STL----unique
    POJ 2031(最小生成树Kruskal算法+几何判断)
    POJ 3468(线段树区间修改+区间求和)
    学习线段树
    POJ 1251(最小生成树裸题)
  • 原文地址:https://www.cnblogs.com/zhangliu/p/7063217.html
Copyright © 2011-2022 走看看