zoukankan      html  css  js  c++  java
  • codeforces285C

    Building Permutation

     CodeForces - 285C 

    Permutation p is an ordered set of integers p1,  p2,  ...,  pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1,  p2,  ...,  pn.

    You have a sequence of integers a1, a2, ..., an. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.

    Input

    The first line contains integer n (1 ≤ n ≤ 3·105) — the size of the sought permutation. The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109).

    Output

    Print a single number — the minimum number of moves.

    Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

    Examples

    Input
    2
    3 0
    Output
    2
    Input
    3
    -1 -1 2
    Output
    6

    Note

    In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).

    In the second sample you need 6 moves to build permutation (1, 3, 2).

    sol:较显然的是最小的变成1,次小的变成2,以此类推,应该是最优的

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    inline ll read()
    {
        ll s=0;
        bool f=0;
        char ch=' ';
        while(!isdigit(ch))
        {
            f|=(ch=='-'); ch=getchar();
        }
        while(isdigit(ch))
        {
            s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
        }
        return (f)?(-s):(s);
    }
    #define R(x) x=read()
    inline void write(ll x)
    {
        if(x<0)
        {
            putchar('-'); x=-x;
        }
        if(x<10)
        {
            putchar(x+'0'); return;
        }
        write(x/10);
        putchar((x%10)+'0');
        return;
    }
    #define W(x) write(x),putchar(' ')
    #define Wl(x) write(x),putchar('
    ')
    const int N=300005;
    int n;
    ll a[N];
    int main()
    {
        int i;
        ll ans=0;
        R(n);
        for(i=1;i<=n;i++) R(a[i]);
        sort(a+1,a+n+1);
        for(i=1;i<=n;i++) ans+=abs(i-a[i]);
        Wl(ans);
        return 0;
    }
    /*
    Input
    2
    3 0
    Output
    2
    
    Input
    3
    -1 -1 2
    Output
    6
    */
    View Code
  • 相关阅读:
    php-fpm 启动参数及重要配置详解
    Linux系统优化实现高并发
    Linux下php-fpm进程过多导致内存耗尽问题
    .htaccess详解及.htaccess参数说明
    添加PHP扩展模块
    nginx 服务器重启命令,关闭
    linux的文件打包与压缩
    vue 存取、设置、清除cookie
    div轮流滚动显示
    滚动监听
  • 原文地址:https://www.cnblogs.com/gaojunonly1/p/10657095.html
Copyright © 2011-2022 走看看