H - Coins

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

//题意是 第一行两个整数 n 和 m (1 <= n <= 100)(m <= 100000) 然后第二行是 n 个硬币的价值，再是 n 个硬币的数量，问这些硬币能组成多少个小于等于 m 的价值。多重背包，有一点难吧，要用到二进制优化，直接当 0 1 背包处理要超时

hud 340ms 另一个oj 187 ms

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 using namespace std;
 5
 6 int a[105],c[105],d[105];
 7 bool f[100005];
 8
 9 int max(int a,int b)
10 {
11     return a>b?a:b;
12 }
13
14 int main()
15 {
16     int i,j,k,m,n;
17     while(scanf("%d%d",&n,&m)&&n+m)
18     {
19         for(i=1;i<=n;i++)
20           scanf("%d",&a[i]);
21         for(i=1;i<=n;i++)
22             scanf("%d",&c[i]);
23
24         memset(f,0,sizeof(f));
25         f[0]=1;
26
27         for(i=1;i<=n;i++)
28         {
29             if(a[i]*c[i]>=m) //如果该货币总价值大于等于 m
30             {
31                 for(j=a[i];j<=m;j++)
32                     if(!f[j])          //只需记得能不能达到即可
33                     f[j]=f[j-a[i]];
34             }
35             else             //总价值小于 m
36             {
37                 if(c[i]==0)
38                     continue;
39
40                 int num=c[i];
41                 int p=1;
42                 int t=1;
43
44                 while(p<num)    //二进制优化的 0 1 背包 1 . 2 . 4 . 8
45                 {
46                     d[t++]=a[i]*p;
47                     num-=p;
48                     p*=2;
49                 }
50
51                 d[t++]=a[i]*num;
52
53                 for(j=1;j<t;j++)
54                 for(k=m;k>=d[j];k--)//f[k]=max(f[k],f[k-d[j]]);
55                 {
56                     if(!f[k])
57                     f[k]=f[k-d[j]];
58                 }
59             }
60         }
61
62         j=0;
63         for(i=1;i<=m;i++)
64         {
65             if(f[i]==1)
66             j++;
67         }
68         printf("%d\n",j);
69     }
70     return 0;
71 }