1.小红与奇数

解题思路：如果给定的数是偶数, 由于1是任意正数的因子, 偶数+1=奇数

若给定的数是奇数, +1/+自身, 都变成了偶数

#include <bits/stdc++.h>
using namespace std;
void solve() {int x;cin >> x;if (x & 1)cout << "No" << '\n';elsecout << "Yes" << '\n';
}
int main() {int t = 1;while (t--) {solve();}
}

 2. 小红与gcd三角形

解题思路：先求gcd(x,y), 然后根据三角形的性质进行判断 

#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {int t = a % b;while (t) {t = a % b;a = b;b = t;}return b;
}
void solve() {int x, y;cin >> x >> y;if (gcd(x, y) > abs(x - y) && gcd(x, y) < x + y) {cout << "Yes" << '\n';} else {cout << "No" << '\n';}
}
int main() {int t;cin >> t;while (t--) {solve();}
}

 3.小红与red

解题思路：如果出现两个连续相同的字符, eg: rrd

中间字符r只要替换为前后都不同的即可

#include <bits/stdc++.h>
using namespace std;
const string t = "red";
void solve() {int n;string s;cin >> n >> s;for (int i = 1; i < n; i++) {if (s[i] == s[i - 1]) {for (char c : t) {if (c != s[i - 1] && (c != s[i + 1] || i == n - 1)) {s[i] = c;break;}}}}cout << s << endl;
}
int main() {int t = 1;while (t--) {solve();}
}

 4. 小红与好数组

回溯即可

dfs函数

参数包括目标和 n，当前路径 p，当前和 sum，以及前一个元素 pre。

终止条件：当当前和等于 n 时，将当前路径加入结果集 ans 中

选择元素：从1到剩余和（n - sum）中选择一个不等于前一个元素的数，继续递归

#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> ans;
void dfs(int n, vector<int>& p, int sum, int pre) {if (sum == n) {ans.push_back(p);return;}for (int cur = 1; cur <= n - sum; cur++) {if (cur != pre) {p.push_back(cur);dfs(n, p, sum + cur, cur);p.pop_back();}}
}
void solve() {int n;cin >> n;vector<int> p;dfs(n, p, 0, -1);sort(ans.begin(), ans.end());for (auto& x : ans) {for (int num : x) {cout << num << " ";}cout << '\n';}
}
int main() {int t = 1;while (t--) {solve();}
}

 5. 小红与gcd和sum

 dp

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e6;
void solve() {int n;cin >> n;vector<ll> a(N + 1, 0);for (int i = 0; i < n; i++) {int val;cin >> val;a[val] += val;}vector<ll> dp(N + 1, 0);for (int g = 1; g <= N; g++) {for (int j = g; j <= N; j += g) {dp[g] += a[j];}}ll max_val = 0;for (int g = 1; g <= N; g++) {if (dp[g] > 0) {max_val = max(max_val, (ll)g * dp[g]);}}cout << max_val << endl;
}
int main() {int t = 1;while (t--) {solve();}return 0;
}

 感谢大家的点赞和关注，你们的支持是我创作的动力！