poj_1742_Coins

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.

题解

明显的多重背包啊，+单调队列的优化。
用pascal做不知怎么回事，总超时。所以用c++。

代码

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define min(x,y) x<y?x:y
#define max(x,y) x>y?x:y
using namespace std;

const int MAXN = 110;
const int MAXD = 100010;

int A[MAXN], C[MAXN];
bool bo[MAXD], f[MAXD];

int main(){
    int n, m;
    while (scanf("%d %d", &n, &m) && n && m >= 0){
        for (int i=1;i<=n;++i)
            scanf("%d",&A[i]);
        for (int i=1;i<=n;++i)
            scanf("%d",&C[i]);
        for (int i=0;i<=m;++i)
            bo[i]=false;
        int num=0; bo[0]=true;
        for (int i=1;i<=n;++i){
            if (C[i]==1){
                for (int j=m;j>=A[i];--j)
                    if (!bo[j] && bo[j-A[i]])
                        bo[j]=true,++num;
                continue;
            }
            if (A[i]*C[i]>=m){
                for (int j=A[i];j<=m;++j)
                    if (bo[j-A[i]] && !bo[j])
                        bo[j]=true,++num;
                continue;
            }
            for (int rem=0;rem<A[i];++rem){
                int s=0,e=-1,sum=0;
                for (int v =rem;v<=m;v+=A[i]){
                    if (s+C[i]==e)
                        sum-=f[s++];
                    f[++e]=bo[v]; sum+=bo[v];
                    if (!bo[v] && sum)
                        bo[v]=true,++num;
                }
            }
        }
        printf("%d
", num);
    }
    return 0;
}