zoukankan      html  css  js  c++  java
  • Codeforces Gym 100463A Crossings 逆序数

    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个数等于(a*i+b)%n,然后问你逆序数是多少

    题解:

    树状数组,大胆上

    代码

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 1000101
    #define mod 10007
    #define eps 1e-9
    const int inf=0x7fffffff;   //无限大
    /*
    inline ll read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    */
    //**************************************************************************************
    int d[maxn];
    int c[maxn];
    ll n;
    int t;
    inline int read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    int lowbit(int x)
    {
        return x&-x;
    }
    
    void update(int x,int y)
    {
        while(x<=n)
        {
            d[x]+=y;
            x+=lowbit(x);
        }
    }
    int sum(int x)
    {
        int s=0;
        while(x>0)
        {
            s+=d[x];
            x-=lowbit(x);
        }
        return s;
    }
    int num[maxn];
    ll a,b;
    int main()
    {
        int t=0;
        while(scanf("%lld%lld%lld",&n,&a,&b)!=EOF)
        {
            t++;
            if(n==0&&a==0&&b==0)
                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
    ",t,(n-1)*n/2-ans);
        }
    }
  • 相关阅读:
    汇编--基础分析:数据段在内存中的存放及空间
    汇编语言(王爽)第六章检测点与实验5
    C语言经典例题100(22~40)
    C:数组小结(3)
    随机不重复的取数组元素,并赋值给div使用
    关于Apple设备私有的apple-touch-icon属性详解
    测试你女友是否跟你结婚插件的源码
    CSS子元素设置margin-top作用于父容器?
    hybrid app开发中:苹果移动设备实用Meta标签
    chromium 修改chromium的设置选项
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4665222.html
Copyright © 2011-2022 走看看