POJ 1017 Packets 贪心 模拟

一步一步模拟，做这种题好累

先放大的的，然后记录剩下的空位有多少，塞1*1和2*2的进去

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#elsefreopen("in.txt","r",stdin);//freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {int ch;while((ch=getchar())!=EOF) {if(ch!=' '&&ch!='\n')return ch;}return EOF;
}
bool run()
{int a[7];for(int i=1;i<=6;i++)cin>>a[i];int cnt=0;for(int i=1;i<=6;i++)if(a[i]!=0)cnt=1;if(!cnt)return 0;int res=0;int d1=0,d2=0;res+=a[6];res+=a[5];d1+=a[5]*11;res+=a[4];d2+=a[4]*5;res+=(a[3]+3)/4;if(a[3]%4!=0){d1+=8-a[3]%4;d2+=7-2*(a[3]%4);}
//    if(a[3]%4==1)
//        d1+=7,d2+=5;
//    else if(a[3]%4==2)
//        d1+=6,d2+=3;
//    else if(a[3]%4==3)
//        d1+=5,d2+=1;
//    写完发现有规律。。int x;x=min(a[2],d2);a[2]-=x;d2-=x;d1+=d2*4;x=min(a[1],d1);a[1]-=x;d1-=x;res+=(a[2]+8)/9;if(a[2]%9!=0)d1+=(9-a[2]%9)*4;x=min(a[1],d1);a[1]-=x;d1-=x;res+=(a[1]+35)/36;printf("%d\n",res);return true;
}
int main()
{std::ios::sync_with_stdio(false);while(run());return 0;
}