POJ 1742 coins (dp)

题目大意是给出一些面值的硬币和数量。问在1-m中能凑出多少种钱。

设 dp[i+1][j] 为前i种凑成j元第i种最多剩下多少。

1. dp[i+1][j] = mi ( dp[i][j]>=0) 前i-1种已经能凑成j了

2.dp[i+1][j] = -1 （ j<a[i] || dp[i+1][j-a[i]] <=0 )

3. dp[i+1][j] = dp[i+1][j-a[i]] -1

题目：

F - Coins

Time Limit:3000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box 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


代码：

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 
 6 int n,m;
 7 int A[200];
 8 int C[200];
 9 int dp[100000+10];
10 int main()
11 {
12     while(scanf("%d%d",&n,&m)!=EOF)
13     {
14         if(n==0)break;
15         for(int i=0;i<=m;i++)
16             dp[i] = -1;
17 
18         for(int i=0;i<n;i++)
19         {
20             scanf("%d",&A[i]);
21         }
22         for(int i=0;i<n;i++)
23         {
24             scanf("%d",&C[i]);
25         }
26         //dp
27         dp[0]=0;
28         for(int i=0;i<n;i++)
29         {
30             for(int j=0;j<=m;j++)
31             {
32                 if( dp[j] >=0)
33                     dp[j] = C[i];
34                 else if( j<A[i] || dp[j-A[i]]<=0)
35                     dp[j] = -1;
36                 else
37                     dp[j] = dp[j-A[i]]-1;
38             }
39         }
40         int cnt =0;
41         for(int i=1;i<=m;i++)
42         {
43             if( dp[i]>=0)cnt++;
44         }
45         printf("%d
",cnt);
46     }
47     return 0;
48 }