给出N种钱币和M

给出N种钱币的面值和个数

NPC拿着这N些钱币去买价值M的物品，能够多付。然后被找零，找零的钱也为这些面值。但没有数量限制

问最少经手的钱币数量

对于NPC做一个付款多重背包

然后对于找零做一个全然背包

 ans=Min(dp1[i]+dp2[i-m],ans);

#include "stdio.h"
#include "string.h"


int n,m;
int dp1[20010],dp2[20010],c[20010],v[20010];
void onezero_pack(int v,int k)
{
    int i;
    for (i=20000;i>=v;i--)
        if (dp1[i-v]!=-1 && (dp1[i-v]+k<dp1[i] || dp1[i]==-1) )
        dp1[i]=dp1[i-v]+k;
}

void complete_pack(int v)
{
    int i;
    for (i=v;i<=20000;i++)
        if (dp1[i-v]!=-1 && (dp1[i-v]+1<dp1[i] || dp1[i]==-1) )
        dp1[i]=dp1[i-v]+1;
}

void multiple_pack(int v,int c)
{
    int k;
    if (v*c>=20000)
        complete_pack(v);
    else
    {
        k=1;
        while (k<c)
        {
            onezero_pack(k*v,k);
            c-=k;
            k*=2;
        }
        if (c>0) onezero_pack(c*v,k);
    }
}

int Min(int a,int b)
{
    if (a<b) return a;
    else return b;
}
int main()
{
    int Case,i,j,ans;
    Case=0;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        if (n+m==0) break;
        for (i=1;i<=n;i++)
            scanf("%d",&v[i]);
        for (i=1;i<=n;i++)
            scanf("%d",&c[i]);

        memset(dp1,-1,sizeof(dp1));
        dp1[0]=0;


        for (i=1;i<=n;i++)
            multiple_pack(v[i],c[i]);

       memset(dp2,-1,sizeof(dp2));
        dp2[0]=0;
        for (i=1;i<=n;i++)
            for (j=0;j<=20000-v[i];j++)
            {
                if (dp2[j]!=-1 && (dp2[j]+1<dp2[j+v[i]] || dp2[j+v[i]]==-1) )
                    dp2[j+v[i]]=dp2[j]+1;
            }

        ans=0x3f3f3f3f;
        for (i=m;i<=20000;i++)
            if (dp1[i]!=-1 && dp2[i-m]!=-1)
                ans=Min(dp1[i]+dp2[i-m],ans);

        printf("Case %d: ",++Case);
        if (ans==0x3f3f3f3f)
            printf("-1\n");
        else
            printf("%d\n",ans);
    }
    return 0;
}