A.只要考虑分成一个串的情况就可以了。
#include<bits/stdc++.h> using namespace std; int n,a[105]; int main() { ios::sync_with_stdio(0); cin >> n; for(int i = 1;i <= n;i++) cin >> a[i]; if(a[1]%2 && a[n]%2 && n%2) cout << "Yes" << endl; else cout << "No" << endl; return 0; }
B.1.若所有点在一条直线,则no。
2.若a[1]单独一组,剩余点在一条直线,则yes。
3.枚举a[1]和后面点组成的直线,判断是否能把点分成两组。
#include<bits/stdc++.h> using namespace std; int n,a[1005]; int main() { ios::sync_with_stdio(0); cin >> n; for(int i = 1;i <= n;i++) cin >> a[i]; int t = a[3]-a[2]; int ok = 0; for(int i = 3;i <= n;i++) { if(a[i]-a[i-1] != t) ok = 1; } if(!ok) { if(a[2]-a[1] == t) cout << "No" << endl; else cout << "Yes" << endl; return 0; } ok = 0; for(int i = 2;i <= n;i++) { double t = 1.0*(a[i]-a[1])/(i-1); vector<int> v; for(int j = 2;j <= n;j++) { if(abs(1.0*(a[j]-a[1])/(j-1)-t) > 1e-8) v.push_back(j); } int flag = 1; for(int j = 1;j < v.size();j++) { if(abs(1.0*(a[v[j]]-a[v[0]])/(v[j]-v[0])-t) > 1e-8) flag = 0; } if(flag) ok = 1; } if(ok) cout << "Yes" << endl; else cout << "No" << endl; return 0; }
C.首先,每个字母相互独立。对于每个字母,最优的方案是一个一个加,打个表,每次加个最大可行量。注意0的特殊情况。
#include<bits/stdc++.h> using namespace std; int n,x[500]; string ans = "z"; int main() { ios::sync_with_stdio(0); x[1] = 0; for(int i = 2;i <= 440;i++) x[i] = x[i-1]+i-1; cin >> n; for(char c = ‘a‘;c <= ‘z‘;c++) { if(n == 0) break; int t = 440; while(x[t] > n) t--; n -= x[t]; while(t--) ans.append(1,c); } cout << ans << endl; return 0; }
D.分成两组,按照g-t排序,这个顺序保证相遇顺序。
之后双指针模拟,遇到连续的几个相同时,画个图便知道那个点从哪里出了。
#include<bits/stdc++.h> using namespace std; int n,w,h,tx[100005],ty[100005],ansx[100005],ansy[100005]; struct xx { int x,t,id; friend bool operator<(xx a,xx b) { if(a.x-a.t != b.x-b.t) return a.x-a.t < b.x-b.t; return a.x < b.x; } }a[100005],b[100005]; int main() { ios::sync_with_stdio(0); cin >> n >> w >> h; int cnt1 = 0,cnt2 = 0; for(int i = 1;i <= n;i++) { int x,y,t; cin >> x >> y >> t; if(x == 1) { a[++cnt1].x = y; a[cnt1].t = t; a[cnt1].id = i; tx[i] = y; ty[i] = h; } else { b[++cnt2].x = y; b[cnt2].t = t; b[cnt2].id = i; tx[i] = w; ty[i] = y; } } sort(a+1,a+1+cnt1); sort(b+1,b+1+cnt2); int now1 = 1,now2 = 1; while(now1 <= cnt1 && now2 <= cnt2) { if(a[now1].x-a[now1].t == b[now2].x-b[now2].t) { int t1 = now1+1,t2 = now2+1; while(t1 <= cnt1 && a[now1].x-a[now1].t == a[t1].x-a[t1].t) t1++; while(t2 <= cnt2 && b[now2].x-b[now2].t == b[t2].x-b[t2].t) t2++; vector<int> v1,v2; for(int i = now1;i < t1;i++) v1.push_back(a[i].id); for(int i = t2-1;i >= now2;i--) v1.push_back(b[i].id); for(int i = t2-1;i >= now2;i--) v2.push_back(b[i].id); for(int i = now1;i < t1;i++) v2.push_back(a[i].id); for(int i = 0;i < v1.size();i++) { ansx[v2[i]] = tx[v1[i]]; ansy[v2[i]] = ty[v1[i]]; } now1 = t1; now2 = t2; } else if(a[now1].x-a[now1].t < b[now2].x-b[now2].t) { ansx[a[now1].id] = tx[a[now1].id]; ansy[a[now1].id] = ty[a[now1].id]; now1++; } else { ansx[b[now2].id] = tx[b[now2].id]; ansy[b[now2].id] = ty[b[now2].id]; now2++; } } while(now1 <= cnt1) { ansx[a[now1].id] = tx[a[now1].id]; ansy[a[now1].id] = ty[a[now1].id]; now1++; } while(now2 <= cnt2) { ansx[b[now2].id] = tx[b[now2].id]; ansy[b[now2].id] = ty[b[now2].id]; now2++; } for(int i = 1;i <= n;i++) cout << ansx[i] << " " << ansy[i] <<endl; return 0; }
Codeforces_849
时间: 2024-10-26 11:22:32