被虐惨了，实验室里数十位大佬中的一位闲来无事切题(二,然后出了5t，当然我要是状态正常也能出5，主要是又热又有蚊子什么的...
题都挺水的。包括F题。
A：
略
B：
找k个最大的数存一下下标然后找段长度就行
public static void main(String[] args) {
int n = nextInt();
int k = nextInt();
PriorityQueue<Integer> q = new PriorityQueue<>();
int a[] = new 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();
}
main 方法
C：
从两个端点向中间遍历
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();
}
D：
大水题，四个位置上的元素判断一下情况就行，我直接分的类很莽就不放代码了。
E：
求出dfs序，顺手存个子树大小，结点下标就完了。
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];//存子树大小
static boolean b[] = new boolean[N];//染色
static int ans[] = new int[N];//dfs序
static int index[] = new int[N];//下标
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];
}
}
}
}
F：很显然上来就想到了从两个点往中间搜搜一半，但是我写炸了，就找了份代码看了看，对于 “一半”，这个地方 要用 (n/2+m/2)而不是 (n+m)/2 ，ex：1 1 10 10.
还是很好懂的 ，哦还需要知道这么一个事， a^b=c 的话 那么 b^c=a; 原谅我是二进制小白，有几个相关公式可以直接百度
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.
*/