hdu2844多重背包简单应用

Coins

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 30   Accepted Submission(s) : 13

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

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

 

Sample Output

84
多重背包的应用，这里边用到了，二进制优化的操作，多重背包，需要完全的理解01背包的基础上深入分析得到的
代码如下
#include<iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;const int maxn=100005,INF=-100000;int a[110],b[110];int f[maxn],n,m;void olbag(int v){    for(int i=m;i>=v;i--)        f[i]=max(f[i],f[i-v]+v);}void completbag(int v){    for(int i=v;i<=m;i++)        f[i]=max(f[i],f[i-v]+v);}void choose(int v,int s){    if(v*s>=m)        completbag(v);    else    {        int k=1;        int cnt=s;        while(k<=cnt)        {            olbag(k*v);            cnt-=k;            k*=2;        }        olbag(cnt*v);    }}int main(){    int counta;    while(scanf("%d %d",&n,&m)!=EOF&&(n||m))    {        counta=0;        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        for(int i=0;i<n;i++)            scanf("%d",&b[i]);        for(int i=0;i<=m;i++)            f[i]=0;        for(int i=0;i<n;i++)        {            choose(a[i],b[i]);        }        for(int i=1;i<=m;i++)            if(f[i]==i)                counta++;        printf("%d\n",counta);    }    return 0;}