E. Essay Time
题目：
分析：
挖坑，后边补
AC代码：
import java.util.*;
import java.io.*;
public class E {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = sc.nextInt();
int z=0;
HashSet<String>used= new HashSet<>();
StringBuffer sol= new StringBuffer();
while (t-->0){
String s = sc.nextLine();
if(s.length()>=4){
if (used.contains(s)){
if (z>0)
sol.append("\n");
z++;
sol.append(s);
}
else used.add(s);
}
}
if (z>0) {
pw.println(z);
pw.println(sol);
}
else pw.println("SAFO");
pw.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public boolean ready() throws IOException {
return br.ready();
}
}
}
G. Grade Concept（线段树）
题目：
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
分析：
线段树问题，维护一个线段树即可，模板问题
AC代码：
import java.io.*;
import java.util.*;
public class G implements Closeable {
private InputReader in = new InputReader(System.in);
private PrintWriter out = new PrintWriter(System.out);
private class Node {
private int idx, lo, hi, min, max, sum;
private void merge(Node left, Node right) {
min = Math.min(left.min, right.min);
max = Math.max(left.max, right.max);
sum = left.sum + right.sum;
}
private void assign(int value) {
min = max = sum = value;
}
}
private class SegmentTree {
private int[] x;
private Node[] nodes;
private SegmentTree(int[] x) {
int n = x.length;
this.x = x;
nodes = new Node[4 * n + 1];
for (int i = 0; i < nodes.length; i++) {
nodes[i] = new Node();
}
init(1, 0, n - 1);
}
private void init(int idx, int left, int right) {
nodes[idx].lo = left;
nodes[idx].hi = right;
if (left == right) {
nodes[idx].assign(x[left]);
} else {
int mid = (left + right) / 2;
init(idx << 1, left, mid);
init(idx << 1 | 1, mid + 1, right);
nodes[idx].merge(nodes[idx << 1], nodes[idx << 1 | 1]);
}
}
private int query(int left, int right) {
Node result = query(1, left, right);
return result.sum - result.min - result.max;
}
private Node query(int idx, int a, int b) {
if (a <= nodes[idx].lo && nodes[idx].hi <= b) {
return nodes[idx];
}
int mid = (nodes[idx].lo + nodes[idx].hi) / 2;
if (a > mid) {
return query(2 * idx + 1, a, b);
}
if (b <= mid) {
return query(2 * idx, a, b);
}
Node result = new Node();
result.merge(query(2 * idx, a, mid), query(2 * idx + 1, mid + 1, b));
return result;
}
private void update(int idx, int value) {
update(1, idx, idx, value);
}
public void update(int idx, int left, int right, int value) {
if (nodes[idx].lo > right || nodes[idx].hi < left) return;
if (nodes[idx].lo == left && nodes[idx].hi == right) {
nodes[idx].assign(value);
return;
}
update(idx << 1, left, right, value);
update(idx << 1 | 1, left, right, value);
nodes[idx].merge(nodes[idx << 1], nodes[idx << 1 | 1]);
}
}
public void solve() {
int n = in.ni(), q = in.ni();
int[] x = new int[n];
for (int i = 0; i < n; i++) {
x[i] = in.ni();
}
SegmentTree tree = new SegmentTree(x);
while (q-- > 0) {
int type = in.ni();
if (type == 1) {
int left = in.ni() - 1, right = in.ni() - 1;
out.println(tree.query(left, right));
} else {
int idx = in.ni() - 1, value = in.ni();
tree.update(idx, value);
}
}
}
@Override
public void close() throws IOException {
in.close();
out.close();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int ni() {
return Integer.parseInt(next());
}
public long nl() {
return Long.parseLong(next());
}
public void close() throws IOException {
reader.close();
}
}
public static void main(String[] args) throws IOException {
try (G instance = new G()) {
instance.solve();
}
}
}
H. High Hopes
题目：
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
分析：
任何数的0次方都等于1.
AC代码：
import java.util.*;
public class h{
public static void main(String[] a) {
Scanner s=new Scanner(System.in);
int n=s.nextInt();
while(n-->0)
System.out.println(0);
}
}