Gym 100463A Crossings （树状数组逆序对）-优快云博客

本文介绍了一个算法问题，即计算特定排列的交叉数。通过定义排列Perm(n, a, b)，文章给出了具体的输入输出示例，并提供了一种有效的解决方法，包括使用逆序对的概念和具体的代码实现。

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. !""""#""""

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

思路：

很明显的逆序对。。照着公式把每个值算出来，然后直接求逆序对个数就好了。

注意要开long long 要不会超时。。。因为中间计算公式的时候值是超过了in范围t的，

不开long long就会陷入死循环

实现代码：

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define ll long long
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 1e6 + 10;
const double EPS = 1e-8;
//inline int sgn(double x) return (x > EPS) - (x < -EPS); //浮点数比较常数优化写法
int c[M],l[M];
ll n;

int lowbit(int x){
    return x&(-x);
}

int getsum(int x){
    int sum = 0;
    while(x>0){
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}

void update(int x,int value){
    while(x<=n){
        c[x] += value;
        x += lowbit(x);
    }
}

int main()
{
    ll a,b;
    int cal = 1;
    while(scanf("%lld%lld%lld",&n,&a,&b)!=EOF){
        memset(c,0,sizeof(c));
        if(n==0&&a==0&&b==0) break;
        for(int i = 1;i <= n;i ++){
            l[i] = ((i-1)*a+b)%n+1;
        }
        ll ans = 0;
        for(int i = 1;i <= n;i ++){
            update(l[i],1);
            ans +=i - getsum(l[i]);
        }

        printf("Case %d: %lld\n",cal++,ans);
    }
    return 0;
}