CSP - S 2020 T1 儒略日

题目链接

大力模拟题

由于我太菜了，考试的时候直接硬生生的从公元前4713年到公园1583年的每一天都给递推出来了，大概有 $2\cdot 10^6$ 天。如果超出了范围就取个模，再特判几下就完事了

结果竟然40min内过了大样例/jk

#include<cstdio>
#include<iostream>
#include<algorithm> 
using namespace std;
const long long Maxn=3600000+10;
const long long d1[]={
    
    0,31,28,31,30,31,30,31,31,30,31,30,31};
const long long d2[]={
    
    0,31,29,31,30,31,30,31,31,30,31,30,31};
struct node{
    
    
	long long x,y,z;
}g[Maxn];
long long n,k,m,q;
inline long long read()
{
    
    
	long long s=0,w=1;
	char ch=getchar();
	while(ch<'0' || ch>'9'){
    
    if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
int main()
{
    
    
//	freopen("julian.in","r",stdin);
//	freopen("julian.out","w",stdout);
//	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
	g[0].x=g[0].y=1;
	g[0].z=4713;
	while(!(g[n].x==31 && g[n].y==12 && g[n].z==1))
	{
    
    
		++n;
		g[n].x=g[n-1].x+1;
		g[n].y=g[n-1].y;
		g[n].z=g[n-1].z;
		long long tmp;
		if(g[n].z%4==1)tmp=d2[g[n].y];
		else tmp=d1[g[n].y];
		if(g[n].x>tmp)++g[n].y,g[n].x=1;
		if(g[n].y>12)g[n].y=1,g[n].z--;
	}
	k=n;
	++n;
	g[n].x=g[n].y=g[n].z=1;
	while(!(g[n].x==4 && g[n].y==10 && g[n].z==1582))
	{
    
    
		++n;
		g[n].x=g[n-1].x+1;
		g[n].y=g[n-1].y;
		g[n].z=g[n-1].z;
		long long tmp;
		if(g[n].z%4==0)tmp=d2[g[n].y];
		else tmp=d1[g[n].y];
//		if(g[n].z%400==0 || (g[n].z%4==0 && g[n].z%100))
//		tmp=d2[g[n].y];
//		else tmp=d1[g[n].y];
		if(g[n].x>tmp)++g[n].y,g[n].x=1;
		if(g[n].y>12)g[n].y=1,g[n].z++;
	}
	++n;
	g[n].x=15,g[n].y=10,g[n].z=1582;
	while(!(g[n].x==31 && g[n].y==12 && g[n].z==1582))
	{
    
    
		++n;
		g[n].x=g[n-1].x+1;
		g[n].y=g[n-1].y;
		g[n].z=g[n-1].z;
		long long tmp;
		if(g[n].z%400==0 || (g[n].z%4==0 && g[n].z%100))
		tmp=d2[g[n].y];
		else tmp=d1[g[n].y];
		if(g[n].x>tmp)++g[n].y,g[n].x=1;
		if(g[n].y>12)g[n].y=1,g[n].z++;
	}
	//1583
	m=n,++n;
	g[n].x=g[n].y=1,g[n].z=1583;
	while(!(g[n].x==31 && g[n].y==12 && g[n].z==1982))
	{
    
    
		++n;
		g[n].x=g[n-1].x+1;
		g[n].y=g[n-1].y;
		g[n].z=g[n-1].z;
		long long tmp;
		if(g[n].z%400==0 || (g[n].z%4==0 && g[n].z%100))
		tmp=d2[g[n].y];
		else tmp=d1[g[n].y];
		if(g[n].x>tmp)++g[n].y,g[n].x=1;
		if(g[n].y>12)g[n].y=1,g[n].z++;
	}
	q=read();
	while(q--)
	{
    
    
		long long tmp=read();
		if(tmp<=k)
		{
    
    
			printf("%lld %lld %lld BC\n",g[tmp].x,g[tmp].y,g[tmp].z);
			continue;
		}
		if(tmp<=m)
		{
    
    
			printf("%lld %lld %lld\n",g[tmp].x,g[tmp].y,g[tmp].z);
			continue;
		}
		// 400 years = 146097 days
		tmp-=m;
		long long cur=tmp/146097ll;
		cur*=400ll;
		tmp%=146097ll;
		tmp+=m;
		printf("%lld %lld %lld\n",g[tmp].x,g[tmp].y,g[tmp].z+cur);
	}
//	printf("%lld %lld %lld\n",k,m,n);
	return 0;
}