CodeForces - 819D,扩展欧几里得

最新推荐文章于 2021-01-02 19:44:34 发布

xing_mo

最新推荐文章于 2021-01-02 19:44:34 发布

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本文详细解析了一种基于数学和算法的石子游戏，玩家轮流取石子，目标是在特定条件下取得最多石子。通过计算各玩家的取石策略，探讨了游戏背后的数学原理和算法实现。

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After studying the beacons Mister B decided to visit alien’s planet, because he learned that they live in a system of flickering star Moon. Moreover, Mister B learned that the star shines once in exactly T seconds. The problem is that the star is yet to be discovered by scientists.

There are n astronomers numerated from 1 to n trying to detect the star. They try to detect the star by sending requests to record the sky for 1 second.

The astronomers send requests in cycle: the i-th astronomer sends a request exactly a**i second after the (i - 1)-th (i.e. if the previous request was sent at moment t, then the next request is sent at moment t + a**i); the 1-st astronomer sends requests a1 seconds later than the n-th. The first astronomer sends his first request at moment 0.

Mister B doesn’t know the first moment the star is going to shine, but it’s obvious that all moments at which the star will shine are determined by the time of its shine moment in the interval [0, T). Moreover, this interval can be split into Tparts of 1 second length each of form [t, t + 1), where t = 0, 1, 2, …, (T - 1).

Mister B wants to know how lucky each astronomer can be in discovering the star first.

For each astronomer compute how many segments of form [t, t + 1) (t = 0, 1, 2, …, (T - 1)) there are in the interval [0, T) so that this astronomer is the first to discover the star if the first shine of the star happens in this time interval.

Input

The first line contains two integers T and n (1 ≤ T ≤ 109, 2 ≤ n ≤ 2·105).

The second line contains n integers a1, a2, …, a**n (1 ≤ a**i ≤ 109).

Output

Print n integers: for each astronomer print the number of time segments describer earlier.

Examples

Input

4 2
2 3

Output

3 1

Input

5 4
1 1 1 1

Output

2 1 1 1

Note

In the first sample test the first astronomer will send requests at moments t1 = 0, 5, 10, …, the second — at moments t2 = 3, 8, 13, … That’s why interval [0, 1) the first astronomer will discover first at moment t1 = 0, [1, 2) — the first astronomer at moment t1 = 5, [2, 3) — the first astronomer at moment t1 = 10, and [3, 4) — the second astronomer at moment t2 = 3.

In the second sample test interval [0, 1) — the first astronomer will discover first, [1, 2) — the second astronomer, [2, 3) — the third astronomer, [3, 4) — the fourth astronomer, [4, 5) — the first astronomer.

题意：相当于有t个石子，分别放在0~t-1这些位置，然后有n个人轮流去取，第一个人第一次取0号石子，之后第i个人取i-1个人后ai号石子，如果该石子已经被取过了就不算，问每个人能取得多少个石子

思路:令 $(\sum_{i=1}^Na_i) mod\ t$ , $t,b1=0b_i = (\sum_{k=2}^ia_i)mod\ t,b_1 = 0$ （表示第i个人第一次取到 $b_i$ 号石子）,则如果第i个人能取到第 $x_i$ 号石子（不一定第一个取到),那么满足下式: $bi+s∗k1≡xi+t∗k2b_i+s*k_1 \equiv x_i+t*k_2$ ,即方程 $s∗k1+t∗k2≡xi−bis*k_1+t*k_2 \equiv x_i-b_i$ （将 $k_2$ 看成 $k_2$ )有解,即对于 $g = g c d (s, t)$ , $xi≡bi(modg)x_i \equiv b_i(mod g)$ ,即对于每个 $b_i mod g$ 相等的人能取到的石子都在一个环上，环上的石子编号满足 $g)x_i \equiv b_i(mod\ g)$ ,对于所有在同一个环上跳跃的人，必经过编号为 $b_i$ %g号石子，设他们从取到的第一个石子跳到该石子所需步数为 $c_i$ ,设任意两个在同一个环上取石子的人i和j，他们跳跃到 $b_i$ %g号石子的步数 $c_i$ 与 $c_j$ 刚好相邻，且cj < ci，那么i能取到的石子数刚好为cj-ci

#include<bits/stdc++.h>
#define MAXN 200005
using namespace std;
int exgcd(int a,int b,int &x,int &y)
{
	if(b == 0)
	{
		y = 0;
		x = 1;
		return a;
	}
	int r = exgcd(b,a%b,x,y);
	int tmp = y;
	y = x - (a/b)*y;
	x = tmp;
	return r;
}
int t,n,a[MAXN],s,ans[MAXN],k1,k2,len,nos[MAXN];
set<int> se[MAXN];
unordered_map<int,int> vis,id;
int main()
{
	while(~scanf("%d%d",&t,&n))
	{
		vis.clear();
		id.clear();
		for(int i = 0;i < MAXN;++i)
			se[i].clear();
		memset(ans,-1,4*n+4);
		a[0] = s = 0;
		for(int i = 1;i <= n;++i)
		{
			scanf("%d",&a[i]);
			s = (s + a[i]) % t;
		}
		a[1] = 0;
		for(int i = 2; i <= n;++i)
			a[i] = (a[i-1] + a[i] ) % t;
		int g = exgcd(s,t,k1,k2),cnt = 0,len = t / g;
		for(int i = 1;i <= n;++i)
		{
			if(!vis.count(a[i]))
			{
				vis[a[i]] = true;
				if(!id[a[i]%g])
					id[a[i]%g] = ++cnt;
				nos[i]=1LL*k1*(a[i]%g-a[i])/g%len;
				nos[i] = (nos[i] + len) % len;
				se[id[a[i]%g]].insert(nos[i]);
			}
			else
				ans[i] = 0;
		}
		
		for(int i = 1;i <= n;++i)
		{
			if(ans[i] == -1)
			{
				set<int>::iterator it = se[id[a[i]%g]].find(nos[i]);
				if(it == se[id[a[i]%g]].begin())
					it = --se[id[a[i]%g]].end();
				else
					--it;
				int s1 = nos[i],s2 = *it,tmp = s1-s2;
				ans[i] = tmp!=0?(tmp+len)%len:len;
			}
		}
		for(int i = 1;i <= n;++i)
			printf("%d ",ans[i]);
		printf("\n");
	}
	return 0;
}