(优化的有限背包) poj 1742 Coins

Coins

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 16723 Accepted: 5807

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

3 101 2 4 2 1 12 51 4 2 10 0

Sample Output

84/*题目大意：略；题目解答：用化成2^k的方法复杂度为O(V*Σlog n[i]),优化后复杂度为o(NV)；ps：楼天成出的题目Source CodeProblem: 1742  User: wawadimu Memory: 920K  Time: 1313MS Language: C++  Result: Accepted Source Code*/ #include<iostream>#include<string>using namespace std;int dp[100001];int A[101];//物品重量int C[101];//物品数目int num[100001];//当前使用第i种物品的数目int n,m;int main(){    int i,j,sum;    while(scanf("%d%d",&n,&m)!=EOF && !(n==0 && m==0))    {        sum=0;        for(i=1;i<=n;i++)            scanf("%d",A+i);        for(i=1;i<=n;i++)            scanf("%d",C+i);        memset(dp,0,sizeof(dp));        dp[0]=1;//背包为空        for(i=1;i<=n;i++)        {            memset(num,0,sizeof(num));            for(j=A[i];j<=m;j++)//多重背包正序计算            {                if(!dp[j] && dp[j-A[i]] && num[j-A[i]] < C[i])//可添加第i种物品，并且没有超出数目限制                {                    dp[j]=1;                    sum++;                    num[j]=num[j-A[i]]+1;//数目增加1                }            }        }        printf("%d/n",sum);    }    return 0;}