Codeforces Round #498 (Div. 3)
阿新 • • 發佈：2018-07-17
public tin lin pub def int() NPU one closed 
被虐慘了，實驗室裏數十位大佬中的一位閑來無事切題(江蘇提高省二,然後出了5t，當然我要是狀態正常也能出5，主要是又熱又有蚊子什麽的...
題都挺水的。包括F題。
A：
略
B：
找k個最大的數存一下下標然後找段長度就行
public static void main(String[] args) { int n = nextInt(); int k = nextInt(); PriorityQueue<Integer> q = new PriorityQueue<>(); int a[] = newmain 方法int[n]; for(int i=0;i<n;i++){ a[i] = nextInt(); q.add(a[i]); } while (q.size()>k){ q.poll(); } boolean b[] = new boolean[n]; int sum = 0; int last = 0; for(int i=n-1;i>=0;i--){ if (q.contains(a[i])){ last=i>last?i:last; sum+=a[i]; b[i] = true; q.remove(a[i]); } } out.println(sum); int temp = -1; for(int i=0;i<last;i++){ if (b[i]){ out.print(i-temp+" "); temp = i; } } out.print(n-1-temp); out.flush(); }
C：
從兩個端點向中間遍歷
public static void main(String[] args) { int n = nextInt(); long d[] = new long[n]; for(int i=0;i<n;i++){ d[i] = nextInt(); } long sum1 = 0; long sum3 = 0; int i=0,j=n-1; long ans = 0; while (j>i){ if (sum3==sum1){ ans = sum1>ans?sum1:ans; sum3+=d[j]; sum1+=d[i]; i++;j--; if (sum3==sum1) ans = sum1>ans?sum1:ans; }else if (sum3<sum1){ sum3+=d[j]; j--; }else if (sum3>sum1){ sum1+=d[i]; i++; } } if (j==i){ if (sum3<sum1){ sum3+=d[j]; }else if (sum3>sum1){ sum1+=d[i]; } if (sum3==sum1) ans = sum1>ans?sum1:ans; } out.print(ans); out.flush(); }View Code
D：
大水題，四個位置上的元素判斷一下情況就行，我直接分的類很莽就不放代碼了。
E：
求出dfs序，順手存個子樹大小，結點下標就完了。
package R498; import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.StringTokenizer; import java.util.Vector; public class Main4 { static BufferedReader in=new BufferedReader(new InputStreamReader(System.in)); static StringTokenizer tok; static boolean hasNext() { while(tok==null||!tok.hasMoreTokens()) try{ tok=new StringTokenizer(in.readLine()); } catch(Exception e){ return false; } return true; } static String next() { hasNext(); return tok.nextToken(); } static long nextLong() { return Long.parseLong(next()); } static int nextInt() { return Integer.parseInt(next()); } static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out)); static int N = 200005; static Vector<Integer> g[] = new Vector[N]; static int size[] = new int[N];//存子樹大小 static boolean b[] = new boolean[N];//染色 static int ans[] = new int[N];//dfs序 static int index[] = new int[N];//下標 static int pos = 0; public static void main(String[] args) { int n = nextInt(); int q = nextInt(); for(int i=1;i<=n;i++){ g[i] = new Vector<>(); } int u; for(int i=2;i<=n;i++){ u = nextInt(); g[u].add(i); } dfs(1); int t,k; while (q--!=0){ t = nextInt(); k = nextInt(); if (k>size[t]){ out.println(-1); }else { out.println(ans[index[t]+k-1]); } } out.flush(); } public static void dfs(int v){ b[v]=true; size[v]++; ans[pos++] = v; index[v] = pos-1; for(int i=0;i<g[v].size();i++){ int u = g[v].get(i); if(!b[u]){ dfs(u); size[v]+=size[u]; } } } }View Code
F：很顯然上來就想到了從兩個點往中間搜搜一半，但是我寫炸了，就找了份代碼看了看，對於 “一半”，這個地方 要用 (n/2+m/2)而不是 (n+m)/2 ，ex：1 1 10 10.
還是很好懂的 ，哦還需要知道這麽一個事， a^b=c 的話 那麽 b^c=a; 原諒我是二進制小白，有幾個相關公式可以直接百度
import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.*; public class Main6 { static BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); static StringTokenizer tok; static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); static boolean hasNext() { while (tok == null || !tok.hasMoreTokens()) try { tok = new StringTokenizer(in.readLine()); } catch (Exception e) { return false; } return true; } static String next() { hasNext(); return tok.nextToken(); } static long nextLong() { return Long.parseLong(next()); } static int nextInt() { return Integer.parseInt(next()); } static int n; static int m; static long k; static long ans = 0; static long a[][]; static Map<Long,Long> map[][]; public static void main(String[] args) { n = nextInt(); m = nextInt(); k = nextLong(); a = new long[n][m]; map = new Map[n][m]; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ a[i][j] = nextLong(); map[i][j] = new HashMap<>(); } } dfsbe(0,0,0); dfsend(n-1,m-1,0); out.print(ans); out.flush(); } public static void dfsbe(int i,int j,long temp){ if (i>=n||j>=m) return; temp ^= a[i][j]; if (i+j==n/2+m/2){ if (map[i][j].get(temp)==null){ map[i][j].put(temp,1l); }else { map[i][j].put(temp,map[i][j].get(temp)+1); } return; } dfsbe(i+1,j,temp); dfsbe(i,j+1,temp); } public static void dfsend(int i,int j,long temp){ if (i<0||j<0) return; if (i+j==n/2+m/2){ long num = temp^k; if (map[i][j].get(num)!=null){ ans+=map[i][j].get(num); } return; } temp^=a[i][j]; dfsend(i-1,j,temp); dfsend(i,j-1,temp); } } /** * 1^1=0 * 0^0=0 * 1^0=1 * 0^1=1 * a ^ b = b ^ a * a ^ b ^ c = a ^ (b ^ c) = (a ^ b) ^ c; * d = a ^ b ^ c 可以推出 a = d ^ b ^ c. * a ^ b ^ a = b. */View Code
Codeforces Round #498 (Div. 3)