标签：晋级 sync 成功 晋级 时间 模拟 timestamp ios def
训练时间：
A - Jin Yong’s Wukong Ranking List (HihoCoder - 1870)
给你n对拓扑关系，找出第一个不符合之前的拓扑关系的拓扑对。
建图，每加入一对拓扑对 x -> y，看看从 y 是否能跑到 x 即可。
#include <bits/stdc++.h> #define FOPI freopen("in.txt", "r", stdin); #define FOPO freopen("out.txt", "w", stdout); using namespace std; typedef long long LL; const int maxn = 2000 + 100; map<string, int> M; string s[maxn], t[maxn]; vector<int> v[maxn]; void build(int x, int y) { v[x].push_back(y); } int n; bool dfs(int x, int y) { for (int i = 0; i < v[x].size(); i++) { if (v[x][i] == y) return false; if (!dfs(v[x][i], y)) return false; } return true; } bool check() { M.clear(); for (int i = 1; i <= n*2; i++) v[i].clear(); int cnt = 0; for (int i = 1; i <= n; i++) { if (!M.count(s[i])) M[s[i]] = ++cnt; if (!M.count(t[i])) M[t[i]] = ++cnt; int x = M[s[i]], y = M[t[i]]; build(x, y); if (!dfs(y, x)) { cout << s[i] << " " << t[i] << endl; return false; } } return true; } int main() { ios::sync_with_stdio(false); cin.tie(0); while(cin >> n) { for (int i = 1; i <= n; i++) cin >> s[i] >> t[i]; if (check()) cout << 0 << endl; } }
B - Heshen‘s Account Book (HihoCoder - 1871)
坑巨多的模拟。幸好场上没开。
#include <bits/stdc++.h> #define FOPI freopen("in.txt", "r", stdin); #define FOPO freopen("out.txt", "w", stdout); using namespace std; typedef long long LL; const int maxn = 200 + 100; string str[maxn]; int l[maxn], tot[maxn]; bool vis[maxn]; vector<LL> ans; bool check(string s) { int len = s.length(); for (int i = 0; i < len; i++) if (!isdigit(s[i])) return false; return true; } bool check_first(string s) { int len = s.length(); int flag = 0; for (int i = len-1; i >= 0; i--) { if (isdigit(s[i])) flag++; else if (s[i] == ‘ ‘) return flag > 0; else return false; } return true; } int main() { ios::sync_with_stdio(false); cin.tie(0); //FOPI; memset(vis, false, sizeof(vis)); int cnt = 0; while(getline(cin, str[++cnt])); cnt--; bool flag = false; int i; for (i = 1; i <= cnt; i++) { string s = str[i]; string last = ""; int len = s.length(); int tmp = i; if (check_first(s)) { int j; for (j = i+1; j <= cnt; j++) if (check(str[j])) s += str[j], vis[j] = true, tmp = cnt; else { tmp = j-1; break; } } len = s.length(); if (i != cnt && isdigit(s[len-1]) && isdigit(str[i+1][0]) && !vis[i+1]) { vis[i+1] = true; for (int j = 0; j < str[i+1].length() && str[i+1][j] != ‘ ‘; j++) s += str[i+1][j]; } stringstream ss; ss.str(s); string a; if (vis[i]) ss >> a; while(ss >> a) { int len = a.length(); if (len > 1 && a[0] == ‘0‘) continue; if (isalpha(a[0]) || isalpha(a[len-1])) continue; LL sum = 0; for (int j = 0; j < len; j++) if (isdigit(a[j])) sum = sum * 10 + a[j] - ‘0‘; ans.push_back(sum), tot[i]++; } i = tmp; if (i == cnt) break; } int sz = ans.size(); for (int i = 0; i < sz-1; i++) printf("%lld ", ans[i]); printf("%lld\n", ans[sz-1]); for (int i = 1; i <= cnt; i++) printf("%d\n", tot[i]); }
I - Palindromes (HihoCoder - 1878)
打个表，找到规律。
然后分类讨论就行了。
#include <bits/stdc++.h> #define FOPI freopen("in.txt", "r", stdin); #define FOPO freopen("out.txt", "w", stdout); using namespace std; const int maxn = 100000 + 1000; int t; char s[maxn], r[maxn]; int main() { scanf("%d", &t); for (int ca = 1; ca <= t; ca++) { scanf("%s", s); int len = strlen(s); if (len == 1) { printf("%c\n", s[0]-1); continue; } if (len == 2 && s[1] == ‘0‘ && s[0] == ‘1‘) { printf("%d\n", 9); continue; } if (s[0] == ‘1‘) { if (s[1] == ‘0‘) { s[1] = ‘9‘; for (int i = 1; i < len; i++) printf("%c", s[i]); for (int i = len-2; i >= 1; i--) printf("%c", s[i]); } else { for (int i = 1; i < len; i++) printf("%c", s[i]); for (int i = len-1; i >= 1; i--) printf("%c", s[i]); } } else { s[0]--; printf("%s", s); for (int i = len-2; i >= 0; i--) printf("%c", s[i]); } puts(""); } }
The 2018 ACM-ICPC Asia Beijing Regional Contest
标签：晋级 sync 成功 晋级 时间 模拟 timestamp ios def
原文地址：https://www.cnblogs.com/ruthank/p/10663823.html
