[AGC003]E

Time limit : 2sec / Memory limit : 256MB

Problem Statement

Snuke got an integer sequence from his mother, as a birthday present. The sequence has (N) elements, and the (i)-th of them is (i). Snuke performs the following (Q) operations on this sequence. The (i)-th operation, described by a parameter (q_i), is as follows:

• Take the first (q_i) elements from the sequence obtained by concatenating infinitely many copy of the current sequence, then replace the current sequence with those (q_i) elements.

After these (Q) operations, find how many times each of the integers (1) through (N) appears in the final sequence.

Constraints

• (1≦N≦10^5)
• (0≦Q≦10^5)
• (1≦qi≦10^18)

All input values are integers.

Input

The input is given from Standard Input in the following format:

(N;Q)
(q_1)
(...)
(q_Q)

Output

Print N lines. The i-th line (1≦(i)≦(N)) should contain the number of times integer (i) appears in the final sequence after the (Q) operations.

Sample Input 1

5 3
6
4
11

Sample Output 1

3
3
3
2
0

Sample Input 2

10 10
9
13
18
8
10
10
9
19
22
27

Sample Output 2

7
4
4
3
3
2
2
2
0
0

题意

有一个数字串S，初始长度为n，是1 2 3 4 …… n。
有m次操作，每次操作给你一个正整数a[i]，你先把S无穷重复，然后把前a[i]截取出来成为新的S。
求m次操作后，每个数字在S中出现的次数。

分析

发现如果(a_i≥a_{i+1})那么(a_i)是无效的，把序列变成严格上升的
然后令f[i]为第i次操作后的序列在最终序列中出现多少次，有f[m]=1
整块整块的直接相乘
如果有边角料，可以发现，长为x边角料与序列前x是相等的
而所有序列除该位不存在以外，前x位都是相等的((a_i)序列严格上升
那就找到一个最大的小于等于x的(a_i)，对f[i]产生贡献，再用同样方式处理边角料
最后x<n存s[x]表示序列1-i在最终序列出现多少次

---

CODE

ps:中途变量混乱，n m num难以分清

#include<cstdio>
#include<iostream>
#define ll long long
using namespace std;
int n,num,m;
ll a[101010];
ll f[101010],s[101010];
void sol(ll x,ll doe)
{
	if (x<a[1])
	{
		s[x]+=doe;
		return;
	}
	int l=1,r=n;
	int y;
	while (l<=r)
	{
		int mid=(l+r)/2;
		if (a[mid]<=x)
		{
			y=mid;
			l=mid+1;
		}
		else r=mid-1;
	}
	f[y]+=doe*(x/a[y]);
	x=x%a[y];
	if (x) sol(x,doe);
}
int main()
{
//	freopen("E.in","r",stdin);
//	freopen("E.out","w",stdout);
	scanf("%d%d",&n,&m);
	num=1;
	a[1]=n;
	ll x;
	for (int i=1;i<=m;++i)
	{
		scanf("%lld",&x);
		while (x<=a[num] && num)
			--num;
		++num;
		a[num]=x;
	}
	m=n;
	n=num;
	f[n]=1;
	for (int i=n-1;i>=1;--i)
	{
		f[i]+=f[i+1]*(a[i+1]/a[i]);
		sol(a[i+1]%a[i],f[i+1]);
	}
	for (int i=a[1];i>=1;--i)
		s[i]+=s[i+1];
	for (int i=1;i<=a[1];++i)
		s[i]+=f[1];
	for (int i=1;i<=m;++i)
		printf("%lld
",s[i]);
	return 0;
}