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留念,,,

Posted by lvweihua at 2017-08-10 12:26:20 on Problem 1017
#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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