#722 (Div. 1) B. Kavi on Pairing Duty（DP）

Posted 2021-11-22 li_wen_zhuo

题目描述

Kavi has 2n points lying on the OX axis, i-th of which is located at x=i.
Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:
Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:
One of the segments A and B lies completely inside the other.
A and B have the same length.
Consider the following example:

A is a good pairing since the red segment lies completely inside the blue segment.
B is a good pairing since the red and the blue segment have the same length.
C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.
Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.
Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.

Input

The single line of the input contains a single integer n (1≤n≤106).

Output

Print the number of good pairings modulo 998244353.

Examples

input
1
output
1
input
2
output
3
input
3
output
6
input
100
output
688750769

Note

The good pairings for the second example are:

In the third example, the good pairings are:

题目大意

有一条直线，上面有2*n个点。对于每对个点对A和B，要满足（下面两条中的一条）：
1、A完全包含B或者B完全包含A
2、A和B的长度相等

题目分析

设：f[i]为点数为2*i的方案数
$假设第一个区间的左端点为1，右端点为x。我们可以来分类讨论：$
$1、x>n时$
$x=n+1，第一个区间外的点数为y=n-1，这些点只能选择条件2，连完这n-1个点之后，剩余点数为0，$$这部分的方案数为f[0]$
$x=n+2，y=n-2，这些点只能选择条件2，连完这n-2个点之后，剩余点数为2，这部分的方案数为f[1]$
$\dots \dots$
$x=2n，第一个区间外的点数为y=0，连完这0个点之后，剩余点数为2\ast n-2，这部分的方案数为f[n-1]$
$我们可以发现，这种情况的答案为\sum _{i=0}^{n-1}f[i]$

$2、x<=n时$
$当连接1和x后，中间有x-2个空位，我们将这x-2个点连接之后，这一块的长度就是x+x-2。也就是说，2\ast n必须要能整除2\ast (x-1)才有解。$
$所以，这种情况的答案为：n因子的个数d[n]。$

$综上，dp的递推式为：f[i]=\sum _{j=0}^{i-1}f[j]+d[i]。$

代码如下

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <algorithm>
#include <iomanip>
#define LL long long
#define ULL unsigned long long
#define PII pair<int,int>
#define PLL pair<LL,LL>
#define PDD pair<double,double>
#define x first
#define y second
using namespace std;
const int N=1e6+5,mod=998244353;
int f[N],s[N],d[N];
int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)				//枚举所有因子i
		for(int j=i;j<=n;j+=i)			//枚举所有i的倍数
			d[j]++;
	
	f[1]=s[1]=1;						//f[]表示方案数，s[]表示f[]的前缀和
	for(int i=2;i<=n;i++)
	{
		f[i]=((LL)s[i-1]+d[i])%mod;
		s[i]=((LL)s[i-1]+f[i])%mod;
	}
	cout<<f[n]<<endl;
	return 0;
}