AcWing 刷题计划：0x00基础算法——AcWing 117. 占卜DIY

1.题目描述：

2.题意：
略。

3.思路：
模拟。 双端队列模拟放卡和收卡，注意的地方就是，只要有4个朝上就可以，以及一定是碰到4个K的时候才退出，即使生命值为空。

4.代码：

//AcWing 117. 占卜DIY
//#include
//#pragma GCC optimize(3,"Ofast","inline")
#include
#include
#include
#include
#include
#include
#include
//#include
#include
#include
#include
#include
#include
#define FAST ios::sync_with_stdio(false)
#define DEV_RND ((int)rand()*RAND_MAX+rand())
#define RND(L,R) (DEV_RND%((R)-(L)+1)+(L))
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i
#define per(i,n,a) for(int i=n-1;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define li inline
#define re register
using namespace std;
//typedef uniform_int_distribution RNDI;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef double db;
typedef long long ll;
typedef long double ld;
const int maxn = 1e5+5;
const int maxm = 100000+5;
const int mod = 1e9+7;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1);
//int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
//li int f(int x){return x==par[x]?par[x]:par[x]=f(par[x]);}
//mt19937 eng(time(0));
//li int RND(int L,int R){RNDI rnd(L,R);return rnd(eng);}
li ll lowbit(ll x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
li ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res%MOD;}
li ll qmul(ll a,ll b,ll MOD=mod){return (a*b-(ll)((long double)a/MOD*b)*MOD+MOD)%MOD;}
li ll inv(ll x,ll p){return qpow(x,p-2,p);}
li ll jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
db f(db x){return x;}
li db sim(db l,db r){return (f(l)+4.*f((l+r)/2.)+f(r))*(r-l)/6.;}
db asr(db l,db r,db ans,db eps){db m=l+(r-l)/2.,L=sim(l,m),R=sim(m,r);return fabs(L+R-ans)<=15.*eps?L+R+(L+R-ans)/15.:asr(l,m,L,eps/2)+asr(m,r,R,eps/2);}
db asr(db l,db r,db eps){return asr(l,r,sim(l,r),eps);}
namespace IO
{
	li ll read()
	{
		ll x=0;char c=getchar();
		while('0'<=c&&c<='9'||c=='A'||c=='J'||c=='Q'||c=='K')
		{
			if(c=='A') x=1;
			else if(c=='J') x=11;
			else if(c=='Q') x=12;
			else if(c=='K') x=13;
			else if(c=='0') x=10;
			else x=c-'0';
			c=getchar();
		}
		return x;
	}
	template<typename T>
	li void write(T x,char t='\n')
	{
		if(x<0){x=-x;putchar('-');};
		static int sta[25];int top=0;
		do{sta[top++]=x%10;}while(x/=10);
		while(top) putchar(sta[--top]+'0');
		putchar(t);
	}
}
using namespace IO;
/*-------------head-------------*/
//
int n,m;
deque<pair<int,bool>> q[14];
li void solve()
{
	rep(i,1,14) rep(j,0,4) q[i].pb(mp(read(),0));
	//rep(i,1,14) {for(auto Q:q[i]) printf("%3d",Q.fi);cout<<"\n";}
	int life=4;
	step:
		auto now=q[13].front();q[13].pop_front();
		while(now.fi==13&&!q[13].empty()){life--;now=q[13].front();q[13].pop_front();}
		if(now.fi!=13) q[now.fi].push_front(mp(now.fi,1));
	while(life)
	{
		auto tmp=now;
		now=q[tmp.fi].back();q[tmp.fi].pop_back();
		if(now.fi==13){life--;if(life)goto step;}
		q[now.fi].push_front(mp(now.fi,1));
	}
	int cnt=0;
	rep(i,1,13)
	{
		int num=0;
		for(auto Q:q[i]) if(Q.se) num++;
		cnt+=num>=4;
	}
	write(cnt);
	//puts("");
}
int main()
{
	//srand(time(0));
	//freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
	solve();
	//for(int QwQ=read();QwQ;QwQ--) solve();
	//while(~scanf("%d",&n)) solve();
	return 0;
}