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留念,,,#include<cstdio>
#include<algorithm>
using namespace std;
int C[7]={0}; //N[1]--N[6]分别代表1-6号物品的个数
int solve(){
int tol=C[6];
tol+=C[5];
C[1]=max(0,C[1]-11*C[5]); //1号尽量多的往5号放置
tol+=C[4];
if(C[2]>=C[4]*5) //2号箱子多
C[2]-=C[4]*5;
else{ //2号箱子不够,用1号箱子补充
C[1]=max(0,C[1]-4*(5*C[4]-C[2]));
C[2]=0;
}
tol+=(C[3]+3)/4; //每4个3号箱子放一个盒子,+3是为了防止小于4的时候,出现除法结果是0的情况
int mod=C[3]%4;
C[3]=0;
if(mod){ //3号由剩余
if(C[2]>=7-2*mod){ //由剩余规律得出的,先找数的规律,然后再进行计算
C[2]-=7-2*mod;
C[1]=max(0,C[1]-(8-mod));
}
else{
C[1]=max(0,C[1]-(36-9*mod-4*C[2])); //必然的,去掉3*3和2*2占用的,剩下的只能由1*1填补
C[2]=0;
}
}
tol+=(C[2]+8)/9; //与C[3]+3的效果一样
mod=C[2]%9;
C[2]=0;
if(mod)
C[1]=max(0,C[1]-(36-4*mod));
tol+=(C[1]+35)/36;
return tol;
}
int main()
{
while(1){
int t=0;
for(int i=1;i<=6;i++){
scanf("%d",&C[i]);
t+=C[i];
}
if(t==0)break;
printf("%d\n",solve());
}
return 0;
}
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