[AGC003]E - Sequential operations on Sequence

2018/01/26 21:45
阅读数 5

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.

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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\n",s[i]);
	return 0;
}
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