D. PolandBall and Polygon
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
PolandBall has such a convex polygon with n veritces that no three of its diagonals intersect at the same point. PolandBall decided to improve it and draw some red segments.
He chose a number k such that gcd(n, k) = 1. Vertices of the polygon are numbered from 1 to n in a clockwise way. PolandBall repeats the following process n times, starting from the vertex 1:
Assume you‘ve ended last operation in vertex x (consider x = 1 if it is the first operation). Draw a new segment from vertex x to k-th next vertex in clockwise direction. This is a vertexx + k or x + k - n depending on which of these is a valid index of polygon‘s vertex.
Your task is to calculate number of polygon‘s sections after each drawing. A section is a clear area inside the polygon bounded with drawn diagonals or the polygon‘s sides.
Input
There are only two numbers in the input: n and k (5 ≤ n ≤ 106, 2 ≤ k ≤ n - 2, gcd(n, k) = 1).
Output
You should print n values separated by spaces. The i-th value should represent number of polygon‘s sections after drawing first i lines.
Examples
input
5 2
output
2 3 5 8 11
input
10 3
output
2 3 4 6 9 12 16 21 26 31
Note
The greatest common divisor (gcd) of two integers a and b is the largest positive integer that divides both a and b without a remainder.
For the first sample testcase, you should output "2 3 5 8 11". Pictures below correspond to situations after drawing lines.
题目大意：一个n边形，从1号点开始，每次走到(x+k-1)%n+1位置，问每次留下来的路径把这个多边形划分成了几个部分。2 ≤ k ≤ n - 2, gcd(n, k) = 1，可以发现一定每个点一定经过一次，插入边的时候对应的点对应了一个区间，两条直线不相交(贡献答案）当且仅当他们的区间无交集，因为一个点只对应一个区间(只考虑出去的那个)，用树状数组维护一下即可。
有个细节：如果2k>n，那么把k转化一下为n-k就可以了。(以为这个细节很显然，然后就没去hack,结果本来房间里10个对的然后就只剩3个没fst，MDZZ)
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<vector>
5 #include<cstdlib>
6 #include<cmath>
7 #include<cstring>
8 using namespace std;
9 #define maxn 1000010
10 #define llg long long
11 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
12 llg n,m,cs[maxn],k,x,y,ans,c[maxn];
13
14 llg lowbit(llg x) {return x&-x;}
15 void add(llg w,llg v)
16 {
17 while (w<=n)
18 {
19 c[w]+=v; w+=lowbit(w);
20 }
21 }
22 llg sum(llg w)
23 {
24 llg ans=0;
25 while (w>0)
26 {
27 ans+=c[w]; w-=lowbit(w);
28 }
29 return ans;
30 }
31
32 llg work(llg x,llg y)
33 {
34
35 llg l=y,r=y+n-k-k;
36 if (r>n)
37 {
38 r-=n;
39 return sum(n)-sum(l-1)+sum(r);
40 }
41 else return sum(r)-sum(l-1);
42 }
43
44 int main()
45 {
46 yyj("D");
47 cin>>n>>k;
48 if (k>n/2) k=n-k;
49 x=1;
50 ans=1;// add(1,1);
51 for (llg i=1;i<=n;i++)
52 {
53 y=x+k;
54 if (y>n) y-=n;
55 ans+=i-work(x,y);
56 add(x,1);
57 x=y;
58 printf("%lld ",ans);
59 }
60 return 0;
61 }