2011 Michigan Invitational Programming Contest

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

给你三个数n,a,b

满足第i个数等于(a*i+b)%n，然后问你逆序数是多少

这个n有1e6呢，所以树状数组和归并都能解决这个题吧

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+10;
int d[N];
ll n,a,b;
void update(int x,int y) {
    while(x<=n) {
        d[x]+=y;
        x+=x&-x;
    }
}
int sum(int x) {
    int s=0;
    while(x>0) {
        s+=d[x];
        x-=x&-x;
    }
    return s;
}
int main() {
    int k=1;
    while(~scanf("%lld%lld%lld",&n,&a,&b)) {
        if(!n&&!a&&!b)
            break;
        memset(d,0,sizeof(d));
        ll ans=0;
        for(int i=0; i<n; i++) {
            int x=(a*i+b)%n+1;
            ans+=sum(x-1);
            update(x,1);
        }
        printf("Case %d: %lld
",k++,(n-1)*n/2-ans);
    }
}