Acwing周赛48（DAY 44)

1：4410. 吃鸡蛋

题目

代码实现

#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int main()
{
    int n,m;
    cin>> n>> m;
    
    int day=0;
    
    while(n)
    {
        day++;
        n--;
        if(day % m==0)
        n++;
        
    }
    cout << day <<endl;
    return 0;
}

2:4411. 三仙归洞

题目

代码实现

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
long n;
int x,a[3]={0,1,2};
int main()
{
    cin >> n>>x;
    n%=6;
    //cout << n <<endl;
    for(int i=0;i<n;i++)
    {
        if(i%2==0)swap(a[0],a[1]);
        else swap(a[1],a[2]);
    }
    cout << a[x];
    return 0;
}

3:构造数组

题目

y总yyds

代码实现

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 200010, MOD = 998244353;

int n;
PII q[N];

int main()
{
    scanf("%d", &n);

    unordered_map<int, int> L(300000), R(300000);
    for (int i = 1; i <= n; i ++ )
    {
        int a;
        scanf("%d", &a);
        R[a] = i;
        if (!L.count(a)) L[a] = i;
    }

    int m = 0;
    for (auto& [k, v]: L) q[m ++ ] = {L[k], R[k]};

    sort(q, q + m);

    int cnt = 0, st = -1, ed = -1;
    for (int i = 0; i < m; i ++ )
        if (q[i].x <= ed) ed = max(ed, q[i].y);
        else
        {
            cnt ++ ;
            st = q[i].x, ed = q[i].y;
        }

    int res = 1;
    for (int i = 0; i < cnt - 1; i ++ )
        res = res * 2 % MOD;

    printf("%d\n", res);
    return 0;}