CF 444B(DZY Loves FFT-时间复杂度)

最新推荐文章于 2021-08-07 06:47:22 发布

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最新推荐文章于 2021-08-07 06:47:22 发布

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分类专栏：时间复杂度

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本文介绍了一个关于快速傅立叶变换的应用挑战，其中涉及到随机序列的生成与特殊卷积运算的计算方法。通过使用FFT来优化计算过程，并提供了一种简单但有效的解题策略。

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DZY loves Fast Fourier Transformation, and he enjoys using it.

Fast Fourier Transformation is an algorithm used to calculate convolution. Specifically, if a, b and c are sequences with length n, which are indexed from 0 to n - 1, and

We can calculate c fast using Fast Fourier Transformation.

DZY made a little change on this formula. Now

To make things easier, a is a permutation of integers from 1 to n, and b is a sequence only containing 0 and 1. Given a and b, DZY needs your help to calculate c.

Because he is naughty, DZY provides a special way to get a and b. What you need is only three integers n, d, x. After getting them, use the code below to generate a and b.

//x is 64-bit variable;
function getNextX() {
    x = (x * 37 + 10007) % 1000000007;
    return x;
}
function initAB() {
    for(i = 0; i < n; i = i + 1){
        a[i] = i + 1;
    }
    for(i = 0; i < n; i = i + 1){
        swap(a[i], a[getNextX() % (i + 1)]);
    }
    for(i = 0; i < n; i = i + 1){
        if (i < d)
            b[i] = 1;
        else
            b[i] = 0;
    }
    for(i = 0; i < n; i = i + 1){
        swap(b[i], b[getNextX() % (i + 1)]);
    }
}

Operation x % y denotes remainder after division x by y. Function swap(x, y) swaps two values x and y.

Input

The only line of input contains three space-separated integers n, d, x (1 ≤ d ≤ n ≤ 100000; 0 ≤ x ≤ 1000000006). Because DZY is naughty, x can't be equal to 27777500.

Output

Output n lines, the i-th line should contain an integer ci - 1.

   Sample test(s)
 

     input
   
3 1 1

     output
   
1
3
2

     input
   
5 4 2

     output
   
2
2
4
5
5

     input
   
5 4 3

     output
   
5
5
5
5
4

Note

In the first sample, a is [1 3 2], b is [1 0 0], so c0 = max(1·1) = 1, c1 = max(1·0, 3·1) = 3, c2 = max(1·0, 3·0, 2·1) = 2.

In the second sample, a is [2 1 4 5 3], b is [1 1 1 0 1].

In the third sample, a is [5 2 1 4 3], b is [1 1 1 1 0].

这题解法‘朴素’得难以置信

转载自http://codeforces.com/blog/entry/12959：

Firstly, you should notice that A, B are given randomly.

Then there're many ways to solve this problem, I just introduce one of them.

This algorithm can get Ci one by one. Firstly, choose an s. Then check if Ci equals to n, n - 1, n - 2... n - s + 1. If none of is the answer, just calculate Ci by brute force.

The excepted time complexity to calculate Ci - 1 is around

where .

Just choose an s to make the formula as small as possible. The worst excepted number of operations is around tens of million.

对于每次询问：

先暴力枚举，看看答案在不在[n-s+1,n]中

否则暴力。

复杂度=O(s+(tot'0'/i)^s*tot'1')

(tot'0'/i)^s表示[n,n-s+1]中没有答案

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
#define MAXX (1000000006+1)
#define N_MAXX (27777500)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
ll n,d,x;
int i,a[MAXN],b[MAXN];
//x is 64-bit variable;
ll getNextX() {
    x = (x * 37 + 10007) % 1000000007;
    return x;
}
void initAB() {
    for(i = 0; i < n; i = i + 1){
        a[i] = i + 1;
    }
    for(i = 0; i < n; i = i + 1){
        swap(a[i], a[getNextX() % (i + 1)]);
    }
    for(i = 0; i < n; i = i + 1){
        if (i < d)
            b[i] = 1;
        else
            b[i] = 0;
    }
    for(i = 0; i < n; i = i + 1){
        swap(b[i], b[getNextX() % (i + 1)]);
    }
}

int q[MAXN]={0},h[MAXN]={0};
int main()
{
//	freopen("FFT.in","r",stdin);
//	freopen("FFT.out","w",stdout);
	
	
	cin>>n>>d>>x;
	initAB();
	
	Rep(i,n) if (b[i]) q[++q[0]]=i;
	Rep(i,n) h[a[i]]=i;
	
//	Rep(i,n) cout<<a[i]<<' ';cout<<endl;
//	Rep(i,n) cout<<b[i]<<' ';cout<<endl;
	
	
	
	Rep(i,n)
	{
		int s=30,ans=0;
		Rep(j,30)
		{
			if (n-j<=0) break;
			int t=h[n-j];
			if (t<=i&&b[i-t]) {ans=n-j; break;} 
		}		
		if (!ans)
		{
			For(j,q[0])
			{
				int t=q[j];
				if (t>i) break;
				ans=max(ans,a[i-t]*b[t]);
			}
		}
		
		printf("%d\n",ans);
		
		
	}
	
	return 0;
}