F. Shift and Reverse

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-08

5.F. Shift and Reverse2023-12-08

6.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

收起

戳这里，看原题

多重思想的巧妙结合

不多说了，看代码就懂了

#include<bits/stdc++.h>
using namespace std;
int up[200006]={0},down[200006]={0};
int a[200006]={0};
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)cin>>a[i];
        for(int i=n+1;i<=2*n;i++)a[i]=a[i-n];//拆”环“成链，真的太巧妙了，隐含着环的感觉

        up[1]=down[1]=1;
        for(int i=2;i<=2*n;i++)
        {
            up[i]=a[i]>=a[i-1]?up[i-1]+1:1;//代表该点在该点所在的最长非单调递减序列中排第几
            down[i]=a[i]<=a[i-1]?down[i-1]+1:1;//同理，不过是最长非单调递增序列
        }
        int ans=2e9;
        for(int i=n;i<=2*n-1;i++)//如果存在，那么序列的末尾一定大于等于n
        {
            int lr=(2*n-i)%n;//left right，意思是右边剩下的数量。特殊点在于当i==n时为零
            if(up[i]==n)ans=min(ans,min(lr,1+n-lr+1));//对于递增序列，要么把右边的数全部往前移，要么反转一下把包括自己在内的左边的数全部往前移（翻转过来之后就成了右边的数）
            if(down[i]==n)ans=min(ans,min(lr+1,1+n-lr));//同理，这里千万要仔细，最好写的时候嘴里念叨念叨
        }
        printf("%d\n",ans==2e9?-1:ans);
        //for(int i=1;i<=2*n;i++)printf("%d  ",a[i]);//debug
        //puts("");
        //for(int i=1;i<=2*n;i++)printf("%d   %d\n",up[i],down[i]);
    }
    return 0;
}