题意：表达能力不好，就不说了。
分析：算是比较简单的背包了，f[i][j][k]表示前i辆车装了j个武器1和k个武器2后还能装武器3的最多个数，再转移即可。
#include <iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))

int f[2][510][510],w[4],s[4],d[4],x[110],y[110],n;

int main()
{
    int i,j,k,now,pre,tem,c1,c2,c3,d4,ca=0,p,q,rew,res;
    int last1,last2,up1,up2;
    while(scanf("%d",&n)&&n)
    {
        if(ca>0) printf("\n");
        for(i=1;i<=3;i++) scanf("%d%d%d",&w[i],&s[i],&d[i]);
        scanf("%d%d%d%d",&c1,&c2,&c3,&d4);
        for(i=1;i<=n;i++) scanf("%d%d",&x[i],&y[i]);
        memset(f,-1,sizeof(f));
        f[0][0][0]=0;
        now=1; pre=0; 
        last1=0; last2=0;
        for(i=1;i<=n;i++)
        {
            up1=last1+min(x[i]/w[1],y[i]/s[1]);  // 加点小优化减少决策范围 从5000ms减到500ms
            if(up1>500) up1=500;
            up2=last2+min(x[i]/w[2],y[i]/s[2]); 
            if(up2>500) up2=500;
            for(j=0;j<=up1;j++)
            {
                for(k=0;k<=up2;k++)
                {
                    f[now][j][k]=-1;
                    for(p=0;p<=j;p++)
                    {
                        if(p*w[1]>x[i]||p*s[1]>y[i]) break;
                        for(q=0;q<=k;q++)
                        {
                            if(p*w[1]+q*w[2]>x[i]||p*s[1]+q*s[2]>y[i]) break;
                            if(f[pre][j-p][k-q]==-1) continue;
                            rew=x[i]-p*w[1]-q*w[2];
                            res=y[i]-p*s[1]-q*s[2];
                            tem=min(rew/w[3],res/s[3]); // 当前这辆车装完p个武器1和q个武器2后还能装武器3的个数
                            if(f[pre][j-p][k-q]+tem>f[now][j][k])
                                f[now][j][k]=f[pre][j-p][k-q]+tem;
                        }
                    }
                    if(f[now][j][k]!=-1)
                    {
                        last1=max(last1,j);
                        last2=max(last2,k);
                    }
                }
            }
            tem=now; now=pre; pre=tem;
        }
        int flag,ans=0,num;
        if(c1*d[1]+c2*d[2]+c3*d[3]>d4) flag=1;  // 选择比较好的组合武器的方式
        else flag=2;
        for(i=0;i<=500;i++)
        {
            for(j=0;j<=500;j++)
            {
                if(f[pre][i][j]==-1) continue;
                k=f[pre][i][j];
                if(flag==1)
                    ans=max(ans,i*d[1]+j*d[2]+k*d[3]);
                else
                {
                    num=min(i/c1,min(j/c2,k/c3));
                    tem=num*d4+(i-num*c1)*d[1]+(j-num*c2)*d[2]+(k-num*c3)*d[3];
                    ans=max(ans,tem);
                }
            }
        }
        printf("Case %d: %d\n",++ca,ans);
    }
    return 0;
}