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hdu

二进制优化多重背包

将硬币的价值看做费用，使用的硬币个数看做价值，将第\(i\)种硬币看成\(c_i\)个价值为\(a_i\)的硬币跑01背包的话时间是\(O(m\sum c)\)的，显然不大行

注意到可以对\(c_i\)直接进行二进制拆分，把它拆成\(log\)个物品（\(2^0,2^1,\cdots,2^k,c_i-2^k(2^{k+1}>c_i)\)），再跑01背包时间就是\(O(m\sum log\ c_i)\)的，可以接受

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 998244353
#define eps 1e-8
int n,m,w[10100],c[10100],val[10100],dp[101000],tot=0;

int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

int main()
{
    while ((scanf("%d%d",&n,&m)) && (n) && (m))
    {
        rep(i,1,n) val[i]=read();tot=0;
        rep(i,1,n)
        {
            int W=read(),tmp=0;
            while (W>=(1<<tmp))
            {
                w[++tot]=(1<<tmp)*val[i];c[tot]=1;
                W-=(1<<tmp);tmp++;
            }
            if (W) {w[++tot]=W*val[i];c[tot]=1;}
        }
        rep(i,1,m) dp[i]=-maxd;
        rep(i,1,tot)
            per(j,m,w[i])
                dp[j]=max(dp[j],dp[j-w[i]]+c[i]);
        int ans=0;
        rep(i,1,m) ans+=(dp[i]>0);
        printf("%d\n",ans);
    }
    return 0;
}