A
/* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back using namespace std; typedef long long ll; const long long mod = 1e9 + 7; const int N = 2e5 + 5; int main() { ll n; cin >> n; if (n == 0) { cout << 0 << endl; return 0; } n++; if (n % 2 == 0) { cout << n / 2 << endl; } else { cout << n << endl; } }
B
/* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back using namespace std; typedef long long ll; const long long mod = 1e9 + 7; const int N = 2e5 + 5; string a, b, c; string str[5]; int num[5][1000]; int ansermaxn[5]; int main() { ll n; cin >> n; for (int i = 1; i <= 3; i++) { cin >> str[i]; } int aim = 0; int ansermax = -1; int ansnow; for (int i = 1; i <= 3; i++) { for (int j = 0; j < str[i].size(); j++) { num[i][str[i][j]]++; } } for (int i = 1; i <= 3; i++) { for (int j = 65; j <= 122; j++) { if (num[i][j] + n <= str[i].size()) { ansnow = num[i][j] + n; if (ansnow > ansermax) { aim = i; ansermax = ansnow; } } else { if (num[i][j] == str[i].size()) { if (n == 1) { ansnow = str[i].size() - 1; } else { ansnow = str[i].size(); } } else { ansnow = str[i].size(); } if (ansnow > ansermax) { aim = i; ansermax = ansnow; } } ansermaxn[i] = max(ansermaxn[i], ansnow); } } int cnt = 0; for (int i = 1; i <= 3; i++) { //cout << ansermaxn[i] << endl; if (ansermaxn[i] == ansermax) { cnt++; } } if (cnt > 1) { cout << "Draw" << endl; return 0; } if (aim == 1) { cout << "Kuro" << endl; } else if (aim == 2) { cout << "Shiro" << endl; } else { cout << "Katie" << endl; } }
C
/* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back using namespace std; typedef long long ll; const long long mod = 1e9 + 7; const int N = 3e5 + 5; vector<int> gra[N]; ll anser = 0; ll reduce = 0; ll number1 = 0; ll number2 = 0; ll number = 0; int flag = 0; void dfs(int x, int pre, int aim, int flag) { if (flag) { number++; } int len = gra[x].size(); for (int i = 0; i < len; i++) { int to = gra[x][i]; if (to == pre) { continue; } if (to == aim) { dfs(to, x, aim, 1); } else { dfs(to, x, aim, flag); } } } int main() { int n; cin >> n; int x, y; int u, v; cin >> x >> y; for (int i = 1; i <= n - 1; i++) { scanf("%d %d", &u, &v); gra[u].pb(v); gra[v].pb(u); } anser = 1LL * (n - 1) * n; dfs(x, -1, y, 0); number1 = number; number = 0; dfs(y, -1, x, 0); number2 = number; anser -= 1LL * number2 * number1; cout << anser << endl; }
D
一开始给你一个空的集合
有两种操作:
①集合中加入一个数字X
②询问给你三个数字Xi Ki Si 查找集合中是否存在数字V满足 ① Ki是Xi和V的因子 ② V<=Si-Ki 当如果有多个满足条件的数输出Xi Xor V最大的那个V
解:
01字典树暴力操作
直接每次insert一个未出现过的数的时候 暴力Insert到每个它的因子里面去
复杂度为:1e5*ln(1e5)*位数(最多为18)=1e7
#include<bits/stdc++.h> using namespace std; const int maxn = 100010; vector<int>G[maxn]; int vis[maxn]; void read(int &x) { x = 0; char c = getchar(); while (c > '9' || c < '0') { c = getchar(); } while (c >= '0' && c <= '9') { x = (x << 3) + (x << 1) + c - '0', c = getchar(); } } struct Trie //01字典树指针版 { struct node { int Min, val; node *ch[2]; node() { Min = maxn; //维护一个最小值 ch[0] = ch[1] = NULL; } }*rt[maxn]; void init() { for (int i = 1; i < maxn; i++) for (int j = i; j < maxn; j += i) //复杂度1e5*ln1e5=1e6 { G[j].push_back(i); //把一个数的因子全部push进去 } for (int i = 1; i < maxn; i++) { rt[i] = new node; //每个数赋予一个新指针 } } void insert(int x) { int Len = G[x].size(); //插入一个数 for (int i = 0; i < Len; i++) { node *cur = rt[G[x][i]]; //暴力枚举insert到这个数的每个因子里面取 cur->Min = min(cur->Min, x); //维护最小值 for (int j = 17; j >= 0; j--) //争取取每位^1的使之异或值最大 { if (cur->ch[x >> j & 1] == NULL) //如果没有的话创造一个新节点 { cur->ch[x >> j & 1] = new node; } cur = cur->ch[x >> j & 1]; cur->Min = min(cur->Min, x); //每个节点维护最小值 } cur->val = x; //最后一个节点的值为x 最后取答案用 } } int query(int x, int k, int s) { if (x % k != 0) //如果k是x和v的gcd的因子的话 x和v都是k的倍数 { return -1; } node *cur = rt[k]; // if (cur->Min > s - x) //如果最小的都不能满足的话 不存在 { return -1; } for (int i = 17; i >= 0; i--) { int tb = x >> i & 1; if (cur->ch[tb ^ 1] != NULL && cur->ch[tb ^ 1]->Min <= s - x) //如果^1的值满足条件 { cur = cur->ch[tb ^ 1]; } else { cur = cur->ch[tb]; } } return cur->val; //返回使之异或值最大的数列值 } } T; int main() { int N, i, j, opt, x, k, s; T.init(); read(N); while (N--) { read(opt); if (opt == 1) { read(x); if (!vis[x]) { vis[x] = 1, T.insert(x); } } else { read(x); read(k); read(s); printf("%d ", T.query(x, k, s)); } } return 0; }