poj1742

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 33998		Accepted: 11554

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

Source

LouTiancheng@POJ

 

poj1742 动态规划 经典多重背包

有n种不同面值的硬币，面值各为A1,A2,A3。。AN，数量各为C1,C2,C3,,,,,CN。给定数m，问这些硬币能组成小于等于m的数中的哪些数，输出这些数的数目。

其实这个vn算法的编码很简单，关键是如何证明这种贪心是正确的，这个比较复杂，，我还在理解中！！！！！！！！

---用单调队列优化的DP已超出了NOIP的范围，

#include<cstdio>
#include<cstring>
using namespace std;
int n,m,ans,w[105],v[105],d[100005],s[100005];
int main()
{
    while(scanf("%d%d",&n,&m)==2&&n&&m){
        int i,j;ans=0;
        for(i=0;i<n;i++) scanf("%d",w+i);
        for(i=0;i<n;i++) scanf("%d",v+i);
        memset(d,0,sizeof(d));
        for(d[i=0]=1;i<n;i++){
            for(j=0;j<=m;j++) s[j]=0;    
            for(j=w[i];j<=m;j++)
                if(!d[j]&&d[j-w[i]]&&s[j-w[i]]<v[i])
                    d[j]=1,s[j]=s[j-w[i]]+1;
        }
        for(i=1;i<=m;i++)
            if(d[i])
                ans++;
        printf("%d
",ans);
    }
    return 0;
}

#include<cstdio>
int n,m,res;
int use[100005],c[105],f[100005];
int main()
{
    f[0]=1;
    while(scanf("%d%d",&n,&m)==2&&n&&m){
        res=0;
        for(int i=1;i<=m;i++) f[i]=0;
        for(int i=0;i<n;i++) scanf("%d",&c[i]);
        for(int i=0,s;i<n;i++){
            for(int j=0;j<=m;j++) use[j]=0;
            scanf("%d",&s);
            for(int j=c[i];j<=m;j++){
                if(!f[j]&&f[j-c[i]]&&use[j-c[i]]<s){//贪心过程
                    use[j]=use[j-c[i]]+1;
                    f[j]=1;
                    res++;
                }
            }
        }
        printf("%d
",res);
    }
    return 0;
}

对比上下两张AC图

太想节省空间，空间没节省了，时间还更长了。

读入完成再处理比边读边处理快！！！