HDOJ 2844

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5307    Accepted Submission(s): 2181

Problem Description

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.

 

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

2009 Multi-University Training Contest 3 - Host by WHU

 

Recommend

gaojie

 

多重背包

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <iomanip>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1|1
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
#define mid ((l + r) >> 1)
#define mk make_pair
const int MAXN = 100 + 50;
const int maxw = 100 + 20;
const int MAXNNODE = 1000 +10;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 10007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
const D e = 2.718281828459;
int a[MAXN], c[MAXN];
int n , m;
int dp[101000];
void ZeroOnePack(int cost)
{
    REPP(i , m , cost)dp[i] = max(dp[i] , dp[i - cost]);
}
void CompletePack(int cost)
{
    FORR(i , cost , m)dp[i] = max(dp[i] , dp[i - cost]);
}
void MultiPack(int cost , int num)
{
    if(num * cost >= m)
    {
        CompletePack(cost);
        return ;
    }
    int k = 1;
    while(k < num)
    {
        ZeroOnePack(cost * k);
        num -= k;
        k << 1;
    }
    ZeroOnePack(num * cost);
}
int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // Online_Judge
    while(scanf("%d%d" , &n , &m) , n||m)
    {
        FOR(i , 0 , n)scanf("%d" , &a[i]);
        FOR(i , 0 , n)scanf("%d" , &c[i]);
        clr(dp , 0);
        dp[0] = 1;
        FOR(i , 0 , n)MultiPack(a[i] , c[i]);
        int ans = 0;
        FORR(i , 1 , m)if(dp[i])ans++;
        printf("%d\n" , ans);
    }
    return 0;
}