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
cpp
#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
java
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
cpp
#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
java
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。
cpp
#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
java
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)
cpp
#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
java
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]);
}
}