Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6279    Accepted Submission(s): 2561

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

题解：英语不好，果然个个都赶脚是坑啊，唉，题意千万读懂了再做题，本题是多重背包的题目，A  表示硬币的价值， C  表示对应硬币的数量；典型的完全背包啊；

     1.首先是 n 表示 n 组数据，第一行输入价值 A， 第二行输入价值对应的数量 C ；用这些价值的硬币，组合出在 1 和 m ，之间的数（包括1，m）；

     2.所以我们可以把，多重背包分成 01 和 完全背包 来解；如果遇见 A[i]*[i]>=m 按照完全背包，否则 01 背包；

AC代码一：

#include
        
         #include
         
          #include
          
           int v[110];int w[110];int dp[100005];using namespace std;int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0&&m==0)            break;        memset(v,0,sizeof(v));        memset(w,0,sizeof(w));        for(int i=1;i<=100005;i++)        dp[i]=-9999999;        dp[0]=1;        for(int i=1; i<=n; i++)            scanf("%d",&v[i]);        for(int i=1; i<=n; i++)            scanf("%d",&w[i]);        for(int i=1; i<=n; i++)        {            if(v[i]*w[i]>=m)//完全背包            {                for(int j=v[i]; j<=m; j++)                {                    dp[j]=max(dp[j],dp[j-v[i]]+v[i]);                }            }            else            {                for(int k=1; k<=w[i]; k=k*2)                {                    for(int j=m; j>=v[i]*k; j--)                    {                        dp[j]=max(dp[j],dp[j-v[i]*k]+v[i]*k);                    }                     w[i]-=k;                }                if(w[i]>0)                {                    for(int j=m; j>=v[i]*w[i]; j--)                        dp[j]=max(dp[j],dp[j-v[i]*w[i]]+v[i]*w[i]);                }            }        }        int count=0;        for(int i=1; i<=m; i++)        {            if(dp[i]>=0)                count++;        }        printf("%d\n",count);    }    return 0;}
          
         
        

AC代码二：

#include
        
           #include
         
           using namespace std;   int a[200],c[200],F[100005];   void inline ZeroOnePack(int ResVal,int ResVol,int BpCap)  {      for(int i=BpCap;i>=ResVol;--i)      {          F[i]=max(F[i],F[i-ResVol]+ResVal);      }  }    void inline CompletePack(int ResVal,int ResVol,int BpCap)  {      for(int i=ResVol;i<=BpCap;++i)      {          F[i]=max(F[i],F[i-ResVol]+ResVal);      }  }    void MultiplePack(int ResVal,int ResVol,int ResNum,int BpCap)  {      if(ResVol*ResNum>=BpCap)     { CompletePack(ResVal,ResVol,BpCap); }      for(int i=0;(1<
          
           <=ResNum;++i)      {          ZeroOnePack((ResVal<
           
            <
            
             >n>>m)      {             if(n+m==0)        break;        memset(F,0,sizeof(F));        for(i=0;i
             
              >a[i]; for(j=0;j
              
               >c[j]; for(i=0;i