B. DZY Loves FFT

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

132

input

5 4 2

output

22455

input

5 4 3

output

55554

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：

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
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              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
              
               =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<
               
                <<' ';cout<
                
                 i) break; ans=max(ans,a[i-t]*b[t]); } } printf("%d\n",ans); } return 0;}
                
               
              
             
            
           
          
         
        
       
      
     

版权声明：本文博主原创文章。博客，未经同意不得转载。