1. 819B Mister B and PR Shifts

大意: 给定排列$p$, 定义排列$p$的特征值为$\sum |p_i-i|$, 可以循环右移任意位, 求最小特征值和对应移动次数.

右移过程中维护增加的个数和减少的个数即可. 

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+50;
int n, a[N];
int dl[N], dr[N];

int main() {
    scanf("%d",&n);
    REP(i,1,n) scanf("%d",a+i);
    ll ret = 0, ans = 0;
    int L = 0, R = 0;
    REP(i,1,n) { 
        ret += abs(a[i]-i);
        if (a[i]>i) { 
            ++L;
            --dl[a[i]-i];
            ++dr[a[i]-i];
        }
        else if (a[i]<=i) { 
            ++R;
            if (a[i]!=1) {
                --dl[n-i+a[i]];
                ++dr[n-i+a[i]];
            }
        }
    }
    ans = ret;
    int pos = 0;
    REP(i,1,n-1) {
        ret += R-L;
        R += dr[i];
        L += dl[i];
        if (a[n-i+1]!=1) --R,++L;
        ret -= abs(a[n-i+1]-n-1);
        ret += abs(a[n-i+1]-1);
        if (ret<ans) pos = i, ans = ret;
    }
    printf("%lld %d\n", ans, pos);
}
View Code

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