D. Array Repetition

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 Triples2023-12-085.F. Shift and Reverse2023-12-086.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. Lights2024-01-17

31.D. Array Repetition2024-01-17

32.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

原题链接

分析

1.如果某次操作之后，数组的长度大于等于k，我们把这样的操作叫最后操作，且最后操作之后的操作都不用考虑，因为不会影响前面的数组，我们把这样
2.最后操作只有两种，一种是加法，一种是乘法
如果是加法，那么k一定等于此时数组的长度，对应的值一定是这次加法加上去的值，也就是末尾值
如果是乘法，那么分为两种一种是数组的长度恰好等于k，那么这次k对应的值也一定是这次操作过后的末尾值
如果k小于数组的长度，那么递归？分治？dp？反正k的值和上一次操作过后的数组的第$kmodlen[now-1]$位等价

构建

记录每次操作后的数组长度和末尾值，循环找出数组长度大于等于k的操作，直到长度等于k,答案就是此时的末尾值

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll last[1000005]={0};
ll len[100005]={0};
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,q;
        cin>>n>>q;
        for(ll i=1;i<=n;i++)
        {
            ll op,x;
            cin>>op>>x;
            if(op==1)
            {
                last[i]=x;
                len[i]=min(len[i-1]+1,(ll)2e18);
            }
            else
            {
                last[i]=last[i-1];
                if((ll)2e18/(x+1)<len[i-1]) len[i]=2e18;//防止溢出
                else len[i]=len[i-1]*(x+1);
            }
        }

        while(q--)
        {
            ll k;
            cin>>k;
            ll times=lower_bound(len+1,len+n+1,k)-len;//如果k点是在某次加法后诞生的，那么一步到位
            while(k!=len[times])
            {
                k=(k-1)%len[times-1]+1;//说明它在乘法之间，回溯等价于在上一步乘法的模数位置
                times=lower_bound(len+1,len+n+1,k)-len;//有点像dp？递归？
            }
            cout<<last[times]<<" ";
        }
        cout<<endl;
    }
    return 0;
}