H - Coins
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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.
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
一维多重背包二进制优化的思考：先看这个问题，100=1+2+4+8+16+32+37，观察可以得出100以内任何一个数都可以由以上7个数选择组合得到，所以对物品数目就不是从0都100遍历，而是0，1，2，4，5，16，32，37遍历，时间大大优化。
hud 340ms 另一个oj 187 ms
1 #include <stdio.h>
2 #include <string.h>
3
4 int a[105],c[105],b[105];
5 bool dp[100005];
6
7 int main()
8 {
9 int n,m,i,j;
10 while (scanf("%d%d",&n,&m)&&n+m)
11 {
12 memset(dp,0,sizeof(dp));
13 dp[0]=1;
14 for (i=1;i<=n;i++)
15 scanf("%d",&a[i]);
16 for (i=1;i<=n;i++)
17 scanf("%d",&c[i]);
18 for (i=1;i<=n;i++)
19 {
20 if (a[i]*c[i]>m)
21 {
22 for (j=a[i];j<=m;j++)
23 if (!dp[j])
24 dp[j]=dp[j-a[i]];
25 }
26 else
27 {
28 if (c[i]==0)
29 continue;
30
31 int num=c[i];
32 int p=1,t=0;
33 while (p<num)//二进制优化成 01 背包问题
34 {
35 num-=p;
36 b[t++]=p*a[i];
37 p*=2;
38 }
39 b[t++]=num*a[i];
40
41 for (p=0;p<t;p++)
42 for (j=m;j>=b[p];j--)
43 if (!dp[j])
44 dp[j]=dp[j-b[p]];
45 }
46 }
47 j=0;
48 for (i=1;i<=m;i++)
49 if (dp[i])
50 j++;
51 printf("%d
",j);
52 }
53 return 0;
54 }