儒略日 题解

考虑二分年份mid,考察mid.1.1距离BC 4713.1.1的距离,剩余的日期我们暴力跳
问题转化为了算mid年到BC 4713年的总天数
这里有一个trick就是我们先把公元前的年份往前平移一年,即把BC 4713到BC 1年,平移到 4712到0年,最后算完年份再减一下,这样计算会方便很多
那么则有 总天数=365×年数(也就是mid+4712)+闰年数量-10×[mid>1582]
对于闰年数量，就不用讲了，都会计算
那就算完了,注意long long以及二分边界即可。

AC代码

#include<bits/stdc++.h>
#define itn int
#define tin int
#define tni int
#define nit int
#define nti int
#define ll long long
using namespace std;
 
int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch>'9')
	{
		if (ch == '-')f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
		x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
	return x * f;
}
ll readll()
{
	ll x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch>'9')
	{
		if (ch == '-')f = -1; ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
		x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
	return x * f;
}
const int dom[] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
int y = -4737, m = 1, d = 1;
int isRUN(int y)
{
	if (y <= 0)
	{
		y = -y;
		return y % 4 == 1;
	}
	if (y <= 1582)
		return y % 4 == 0;
	return (y % 400 == 0) || ((y % 4 == 0) && (y % 100 != 0));
}
void getnxtday()
{
	d++;
	if (y == 1582 && m == 10 && d == 5)d = 15;
	if (d > dom[m] + (m == 2 && isRUN(y)))m++, d = 1;
	if (m > 12)y++, m = 1;
}
ll checkday(int x)
{
	ll day = 1ll * (x + 4712) * 365;
	if (x > 1582)
		day -= 10;
	day += 1ll * (x + 4712 + 3) / 4;
	if (x >= 1600)
	{
		x -= 1601;
		day -= (ll)x / 100ll;
		day += (ll)x / 400ll;
	}
	return day;
}
int main()
{
	int T;
	T = read();
	while (T--)
	{
		y = 0; m = d = 1;
		ll n = readll();
		int l = -4712, r = 1e9 + 1;
		while (l <= r)
		{
			int mid = (l + r) / 2;
			if (checkday(mid) <= n)
			{
				y = mid;
				l = mid + 1;
			}
			else r = mid - 1;
		}
		n -= checkday(y);
		if (y <= 0)y--;
		while (n--)
			getnxtday();
		printf("%d %d ", d, m);
		if (y <= 0)
			printf("%d BC\n", -y);
		else
			printf("%d\n", y);
	}
	return 0;
}

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