A
B
C
题目给你一个结论 最少需要min((odd,even)个结点可以把一棵树的全部边连起来 要求你输出两颗树
一棵树结论是正确的 另外一棵结论是正确的 正确结论的树很好造 主要是错误的树
题目给了你提示 提供了一个八个结点的错误的树 然后我们慢慢推发现只要N>=6就存在错误的树(把提供的树的左边两个结点删掉)
结点大于6就全部放在4号结点下
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation map<string, ll> mp; string str[100005]; queue<int> que; int main() { int n; cin >> n; if (n < 6) { cout << -1 << endl; } else { if (n % 2) { cout << 1 << " " << 2 << endl; cout << 2 << " " << 3 << endl; cout << 2 << " " << 4 << endl; cout << 4 << " " << 5 << endl; cout << 4 << " " << 6 << endl; cout << 4 << " " << n << endl; for (int i = 7; i <= n - 1; i++) { if (i % 2) { cout << 1 << " " << i << endl; } else { cout << 2 << " " << i << endl; } } } else { cout << 1 << " " << 2 << endl; cout << 2 << " " << 3 << endl; cout << 2 << " " << 4 << endl; cout << 4 << " " << 5 << endl; cout << 4 << " " << 6 << endl; for (int i = 7; i <= n; i++) { if (i % 2) { cout << 1 << " " << i << endl; } else { cout << 2 << " " << i << endl; } } } } for (int i = 1; i <= n - 1; i++) { cout << i << " " << i + 1 << endl; } }
D
玄学暴力题
给你一个数列 要求你给出字典序最小的但不小于给定数列的目标数列 要求目标数列内两两互质
假设我们要求出这个数列可能要求的最大的数 质数的数量级是x/logx 所以 x/logx-1e4>1e5 大概可以求出x在2e6差不多
然后把2-2e6的每个数都存到一个set里面这个set存的是当前所有可插入原数组的数 同时把每个数的质因数都存到一个vector里面
然后输入原有的数组 每次输入一个数就在set里去除掉他的质因数的倍数(包括它本身) 这样这个set里面的每个数就都是当前合法插入数
如果需要插入的数比原数列的大 就可以直接输出set.begin()
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation bool prime[2200005]; vector<int> beishu[2200005]; bool eras[2200005]; set<int> num; bool pre = true; int main() { int n; cin >> n; for (int i = 2; i <= 2200000; i++) { num.insert(i); if (prime[i]) { continue; } //cout<<i<<endl; for (int j = i; j <= 2200000; j += i) { prime[j] = true; beishu[j].pb(i); } } //TS; int now; int aim; for (int i = 1; i <= n; i++) { scanf("%d", &now); if (pre) { aim = *num.lower_bound(now); if (aim > now) { pre = false; } } else { aim = *num.begin(); } cout << aim << " "; for (int j : beishu[aim]) { for (int k = j; k < 2200005; k += j) { if (!eras[k]) { num.erase(k); eras[k] = true; } } } } return 0; }
E
找规律或者OEIS
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll dp[3000005]; ll dfs(ll x) { if (x <= 2000000) { return dp[x]; } if (x % 2) { return 2LL * dfs(x / 2) + x / 2 + 1; } else { return 2LL * dfs(x / 2) + x / 2; } } int main() { ll n; cin >> n; ll anser; dp[0] = 0; for (int i = 0; i <= 1000000; i++) { dp[i * 2] = 2 * dp[i] + i; dp[i * 2 + 1] = 2 * dp[i] + i + 1; } //cout << dp[n - 1] << endl; // for(int i=1;i<=10;i++) // cout<<dp[i]<<endl; cout << dfs(n - 1) << endl; return 0; }