Crossings

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100463

Description

Given a permutation P of {0, 1, ..., n − 1}, we define the crossing number of it as follows. Write the sequence 0, 1, 2, . . . , n − 1 from left to right above the sequence P(0), P(1), . . . , P(n − 1). Draw a straignt line from 0 in the top line to 0 in the bottom line, from 1 to 1, and so on. The crossing number of P is the number of pairs of lines that cross. For example, if n = 5 and P = [1, 3, 0, 2, 4], then the crossing number of P is 3, as shown in the figure below. !""""#""""""""""""&" In this problem a permutation will be specified by a tuple (n, a, b), where n is a prime and a and b are integers (1 ≤ a ≤ n − 1 and 0 ≤ b ≤ n − 1). We call this permutation Perm(n, a, b), and the ith element of it is a ∗ i + b mod n (with i in the range [0, n − 1]). So the example above is specified by Perm(5, 2, 1).

Input

There are several test cases in the input file. Each test case is specified by three space-separated numbers n, a, and b on a line. The prime n will be at most 1,000,000. The input is terminated with a line containing three zeros.

Output

For each case in the input print out the case number followed by the crossing number of the permutation. Follow the format in the example output.

Sample Input

5 2 1 19 12 7 0 0 0

Sample Output

Case 1: 3 Case 2: 77

HINT

题意

给你n个数，   第i-1个数等于(a*i+b)%n，问逆序对有多少

题解：

树状数组

代码

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <queue>
10 #include <map>
11 #include <stack>
12 #define MOD 1000000007
13 #define maxn 32001
14 using namespace std;
15 typedef __int64 ll;
16 inline ll read()
17 {
18     ll x=0,f=1;
19     char ch=getchar();
20     while(ch<‘0‘||ch>‘9‘)
21     {
22         if(ch==‘-‘)f=-1;
23         ch=getchar();
24     }
25     while(ch>=‘0‘&&ch<=‘9‘)
26     {
27         x=x*10+ch-‘0‘;
28         ch=getchar();
29     }
30     return x*f;
31 }
32 //*******************************************************************
33
34 struct ss
35 {
36     int v,index;
37 } in[1000005];
38 int c[1000005];
39 int a[1000005];
40 ll n;
41 bool cmp(ss s1,ss s2)
42 {
43     return s1.v<s2.v;
44 }
45 int lowbit(int x)
46 {
47     return x&(-x);
48 }
49 int getsum(int x)
50 {
51     int sum=0;
52     while(x>0)
53     {
54         sum+=c[x];
55         x-=lowbit(x);
56     }
57     return sum;
58 }
59 void update(int x,int value)
60 {
61     while(x<=n)
62     {
63         c[x]+=value;
64         x+=lowbit(x);
65     }
66 }
67
68 int main()
69 {
70     ll x,y;
71     int oo=1;
72     while(scanf("%lld%lld%lld",&n,&x,&y)!=EOF)
73     {
74         if(n==0&&x==0&&y==0)break;
75         memset(c,0,sizeof(c));
76         for(int i=0; i<n; i++)
77         {
78             a[i+1]=(x*i+y)%n+1;
79         }
80         ll ans=0;
81         for(int i=1; i<=n; i++)
82         {
83             update(a[i],1);
84             ans+=i-getsum(a[i]);
85         }
86         printf("Case %d: ",oo++);
87         cout<<ans<<endl;
88     }
89
90
91     return 0;
92 }