牛客周赛 Round 18_游游拿到一个正整数x,她希望把这个数的前k位进行翻转

牛客竞赛_ACM/NOI/CSP/CCPC/ICPC算法编程高难度练习赛_牛客竞赛OJ (nowcoder.com)

A-游游的整数翻转_牛客周赛 Round 18 (nowcoder.com)

题目描述 

游游拿到了一个正整数�x，她希望把这个整数的前�k位进行翻转。你能帮帮她吗？

简单的字符串操作

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int k;
    string s;
    cin >> s >> k;
    int n = s.size();
    string t = s.substr(0,k);
    reverse(t.begin(),t.end());
    while(t.size() && t[0] == '0')t.erase(0,1);
    string ans = t + s.substr(k,n);
    cout << ans << endl;
    return 0;
}

Java版

import java.util.Scanner;
public class Main 
{
    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int k = sc.nextInt();
        int n = s.length();
        String t = s.substring(0, k);
        t = new StringBuilder(t).reverse().toString();
 
        while (t.length() > 0 && t.charAt(0) == '0')
        t = t.substring(1);
 
        String ans = t + s.substring(k, n);
        System.out.println(ans);
    }
}

 B-游游的排列统计_牛客周赛 Round 18 (nowcoder.com)

游游想知道，有多少个长度为nnn的排列满足任意两个相邻元素之和都不是素数。你能帮帮她吗？

我们定义，长度为nnn的排列值一个长度为nnn的数组，其中1到nnn每个元素恰好出现了一次。

小数素数判断 + 全排列即可

#include<bits/stdc++.h>
using namespace std;
bool No_prime[20];
void Init()
{
    for(int i = 3; i < 20; i++)
    {
        for(int j = 2;j < i;j++)
        if(!(i % j))No_prime[i] = true;
    }
}
int a[11];
int main()
{
    Init();
    int n;cin >> n;
    int ans = 0;
    for(int i = 1;i <= 10;i++)a[i] = i;
    do
    {
        bool is = true;
        for(int i = 1;i < n;i++)
        {
            if(false == No_prime[a[i] + a[i + 1]])
            {
                is = false;
                break;
            }
        }
        ans += is;
    }while(next_permutation(a + 1,a + 1 + n));
    cout << ans << endl;
    return 0;
}

Java版

import java.util.*;
 
public class Main 
{
    static boolean[] No_prime = new boolean[20];
    static int[] a = new int[11];
    public static void init() 
    {
        for (int i = 3; i < 20; i++) 
        for (int j = 2; j < i; j++) 
        if (i % j == 0) 
        {
            No_prime[i] = true;
            break;
        }
    }
 
    public static void main(String[] args) 
    {
        init();
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int ans = 0;
        for (int i = 1; i <= 10; i++)a[i] = i;
        do 
        {
            boolean is = true;
            for (int i = 1; i < n; i++) 
                if (!No_prime[a[i] + a[i + 1]]) 
                {
                    is = false;
                    break;
                }
            ans += is ? 1 : 0;
        } while (next_permutation(a, 1, 1 + n));
        System.out.println(ans);
    }
 
    public static boolean next_permutation(int[] a, int l, int r) 
    {
        int i = r - 1;
        while (i > l && a[i - 1] >= a[i])i--;
        if (i <= l)return false;
        int j = r - 1;
        while (a[j] <= a[i - 1])j--;
        int temp = a[i - 1];
        a[i - 1] = a[j];
        a[j] = temp;
        j = r - 1;
        while (i < j) 
        {
            temp = a[i];
            a[i] = a[j];
            a[j] = temp;
            i++;
            j--;
        }
        return true;
    }
}

 C-游游刷题_牛客周赛 Round 18 (nowcoder.com)

游游制定了一个刷题计划，她找到了nnn套试卷，每套试卷的题目数量为aia_iai​。游游每天上午最多打开一套试卷，下午最多打开一套试卷，也可以选择不刷题而摸鱼。当游游打开一套试卷后，她就会把上面的题目全部刷完。但是游游有强迫症，她希望每天刷的题目总数均为kkk的倍数。请你计算游游最多能刷多少天的题？ 

思路类似力扣中的俩数之和,很好理解。 

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int a[maxn];
int main()
{
    int ans = 0;
    int n, k;cin >> n >> k;
    map<int,int>M;
    for(int i = 1;i <= n;i++)
    {
        int m;cin >> m;
        if(m % k == 0)ans++;
        else
        {
            int p = m % k;
            if(M[k - p] > 0)
            {
                ans++;
                M[k - p]--;
            }
            else
            M[m % k]++;
        }
    }
    cout << ans << endl;
    return 0;
}

 Java版

import java.util.*;
public class Main
{
    public static void main(String[] args) 
    {
        Scanner scanner = new Scanner(System.in);
        int ans = 0;
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        Map<Integer, Integer> M = new HashMap<>();
        for (int i = 1; i <= n; i++) 
        {
            int m = scanner.nextInt();
            if (m % k == 0)ans++;
            else 
            {
                int p = m % k;
                if (M.containsKey(k - p) && M.get(k - p) > 0) 
                {
                    ans++;
                    M.put(k - p, M.get(k - p) - 1);
                } 
                else M.put(m % k, M.getOrDefault(m % k, 0) + 1); 
            }
        }
        System.out.println(ans);
    }
}

D-游游买商品_牛客周赛 Round 18 (nowcoder.com)

背包问题变形(太久没写,晚上看的时候细节容易写错)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1010;
#define endl '\n'
#define int long long
int a[maxn], b[maxn];
int dp[maxn][maxn];//前i件商品价格为j的最大满足
int n, x;
 
signed main()
{
    cin.tie(0) -> sync_with_stdio(false);
    int n, x; cin >> n >> x;
    for(int i = 1;i <= n; i++) cin >> a[i];
    for(int i = 1;i <= n; i++) cin >> b[i];
    for(int i = 1;i <= n; i++)
    {
        for(int j = 1;j <= x; j++)
        {
            if(j >= a[i])
            dp[i][j] = max(dp[i - 1][j],dp[i - 1][j - a[i]] + b[i]);
            else dp[i][j] = dp[i - 1][j]; 
            int t = (a[i] / 2 + a[i - 1]);
            if(i > 1 && j >= t)
            dp[i][j] = max(dp[i][j], dp[i - 2][j - t] + b[i] + b[i - 1]);
        }
    }
    cout << dp[n][x] << endl;
    return 0;
}

Java版

import java.util.*;
 
public class Main 
{
    public static void main(String[] args) 
    {
        Scanner scanner = new Scanner(System.in);
        int maxn = 1010;
        long [][] dp = new long  [maxn][maxn];
        int n, x;
        n = scanner.nextInt();
        x = scanner.nextInt();
        int[] a = new int[maxn];
        int[] b = new int[maxn];
        for (int i = 1; i <= n; i++) a[i] = scanner.nextInt();
        for (int i = 1; i <= n; i++) b[i] = scanner.nextInt();
        for (int i = 1; i <= n; i++) 
        {
            for (int j = 1; j <= x; j++)
            {
                if (j >= a[i]) 
                dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - a[i]] + b[i]);
                else 
                dp[i][j] = dp[i - 1][j];
                int t = (a[i] / 2 + a[i - 1]);
                if (i > 1 && j >= t) 
                dp[i][j] = Math.max(dp[i][j], dp[i - 2][j - t] + b[i] + b[i - 1]);
            }
        }
        System.out.println(dp[n][x]);
    }
}