SDAU 课程练习3 1020

Problem T

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 2

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.<br><br>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

3 101 2 4 2 1 12 51 4 2 10 0

 

Sample Output

84

 

题目大意：

现在有好多硬币，面值分别为   a1.a2....ai，给出每个硬币的数量，求可以组成多少个不同的价值。

思路：

一开始的思路还是将相同的物品转化成多件物品，然后进行  0  1  求解，不出意料的超时。后来继续看了看背包九讲，明白了这个是多重背包的问题。

关于多重背包求解，具体的思路是这样。

当这件物品的数量*这件物品的体积  >=    背包容量   那么这件这就相当于完全背包，完全背包的定义：每件物品无限可用。

如果不够的话，那么就用  0  1  来解，把  1,2,3...k  件物品当成一件物品来看。

然后就这样喽。

感想：

。。。。。。。什么什么的好麻烦。最近心态不行。很无奈。

AC代码：

#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream>using namespace std;int val[900005];int num[900005];int dp[900005];int v;void ZeroOnePack(int cost,int weight){    for(int i=v;i>=cost;i--)       if(dp[i-cost]+weight>dp[i]) dp[i]=dp[i-cost]+weight;}void CompletePack(int cost,int weight){    for(int i=cost;i<=v;i++)        if(dp[i-cost]+weight>dp[i]) dp[i]=dp[i-cost]+weight;}void MultiplePack(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<<=1;        }        ZeroOnePack(amount*cost,amount*weight);    }}int main(){    //freopen("r.txt","r",stdin);    int i,n,m,j;    while(~scanf("%d%d",&n,&v))    {        if(n==0&&v==0) break;        for(i=0;i<n;i++)        {            scanf("%d",&val[i]);        }        for(i=0;i<n;i++)        {            scanf("%d",&num[i]);        }        memset(dp,0,sizeof(dp));        for(i=0;i<n;i++)        {            MultiplePack(val[i],val[i],num[i]);        }        int temp=0,cou=0;        for(i=0;i<=v;i++)        {            if(temp!=dp[i])            {                temp=dp[i];                cou++;            }        }        cout<<cou<<endl;    }    return 0;}