I. Rise of Shadows

Posted 2021-10-24 goto_1600

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题意：给定H和M,每天有H小时M分钟，求时针和分针角度<=$\alpha$的整分针数。
思路：
$\alpha =\frac{2\pi A}{HM}$，可以发现每分钟时针的速率是$v_{时}=\frac{2\pi }{HM}$，分针是$v_{分}=\frac{2\pi }{M}$,他俩的差值是$\frac{2\pi (H-1)}{HM}$我们会发现两个指针一块动的时候不太好计算，我们尝试固定时针，那么现在就变成了追及相遇问题,也就是
$$\begin{gathered} t\ast (H-1)\%(HM)<=|A| \\ -A<=t\ast (H-1)\%(HM)<=At\in [0,H\ast M-1] \\ 两边同时除以g \\ -\frac{A}{g}<=t\ast \frac{H-1}{g}<=\frac{A}{g}\%(\frac{HM}{g}) \\ 由于同余的性质，a\ast c\equiv b\ast c(\%m) \\ 等价于a\equiv b(\%(\frac{m}{gcd(c,m)}) \\ 我们先算正数的，算出正数以后负数是对称的 \\ t<=\frac{A}{g}(\%(H\ast M/g))g=\gcd (HM,H-1)t\in [0,H\ast M/g-1] \\ 那么这个数量就是2\ast \frac{A}{g}+1,这里的1就是0点的贡献 \end{gathered}$$
AC代码：

//#pragma GCC target("avx")
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast")
// created by myq
#include<iostream>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cctype>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<set>
#include<map>
#include<sstream>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N = 400010;
const int mod=998244353;
inline int read()
{
	int res=0;
	int f=1;
	char c=getchar();
	while(c>'9' ||c<'0')
	{
		if(c=='-')	f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		res=(res<<3)+(res<<1)+c-'0';
	}
	return res;
 }
#define int long long
signed main()
{
	int h,m;
	cin>>h>>m;
	int a;
		cin>>a;
	int g=__gcd(h-1,h*m);
	cout<<min(h*m,g*(1+2*(a/g)));
	return 0;

}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/