D - Popcount and XOR
easy.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
int a = rd.nextInt();
int b = rd.nextInt();
long c = rd.nextLong();
int n = 0;
int[] arr = new int[60];
//calculate n,l;
for (int i = 59; i >= 0; i--) {
if (((1L << i) & c) > 0) {
n++;
arr[i] = 1;
}
}
//取k個數k[0,min(n,a)];
//a-k<=60-n;k>=a+n-60;
//2k=a+n-b; 有解
//
if ((a + n - b) < 0 || (a + n - b) % 2 != 0) {
System.out.println(-1);
return;
}
int k = (a + n - b) / 2;
if (k > a || k > n || k < a + n - 60) {
System.out.println(-1);
return;
}
long x = 0;
long y = 0;
int cnt = 0;
for (int i = 0; i < 60; i++) {
if ( arr[i] == 0) {
continue;
}
else if (cnt < k) {
x += (1L<< i);
cnt++;
}
else {
y += (1L<<i);
}
}
//x
for (int i = 0; i < 60; i++) {
if ( arr[i] == 0 && cnt < a) {
x += (1L<<i);
cnt++;
y += (1L<<i);
}
}
System.out.println(x +" " + y);
}
}
class rd {
static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer tokenizer = new StringTokenizer("");
// nextLine()讀取字串
static String nextLine() throws IOException {
return reader.readLine();
}
// next()讀取字串
static String next() throws IOException {
while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(reader.readLine());
return tokenizer.nextToken();
}
// 讀取一個int型數值
static int nextInt() throws IOException {
return Integer.parseInt(next());
}
// 讀取一個double型數值
static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
// 讀取一個long型數值
static long nextLong() throws IOException {
return Long.parseLong(next());
}
// 讀取一個BigInteger
static BigInteger nextBigInteger() throws IOException {
BigInteger d = new BigInteger(rd.nextLine());
return d;
}
}
E - Set Add Query
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
int n = rd.nextInt();
int q = rd.nextInt();
//字首和
long[] sum = new long[q + 1];
long[] nsum = new long[n + 1];
//idx map;
Map<Integer, Integer> map = new HashMap<>();
for (int i = 1; i <= q; i++) {
sum[i] = sum[i - 1];
int x = rd.nextInt();
if (map.containsKey(x)) {
int j = map.get(x);
nsum[x] += sum[i] - sum[j - 1];
map.remove(x);
sum[i] += map.size();
} else {
map.put(x, i);
sum[i] += map.size();
}
}
//再計算沒有closed；
for (Map.Entry<Integer, Integer> entrys : map.entrySet()) {
int x = entrys.getKey();
int id = entrys.getValue();
nsum[x] += sum[q] - sum[id - 1];
}
PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
for (int i = 1; i <= n; i++) {
if (i != 1) {
pw.print(" ");
}
pw.print(nsum[i]);
}
pw.println();
pw.flush();
}
}
class rd {
static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer tokenizer = new StringTokenizer("");
// nextLine()讀取字串
static String nextLine() throws IOException {
return reader.readLine();
}
// next()讀取字串
static String next() throws IOException {
while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(reader.readLine());
return tokenizer.nextToken();
}
// 讀取一個int型數值
static int nextInt() throws IOException {
return Integer.parseInt(next());
}
// 讀取一個double型數值
static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
// 讀取一個long型數值
static long nextLong() throws IOException {
return Long.parseLong(next());
}
// 讀取一個BigInteger
static BigInteger nextBigInteger() throws IOException {
BigInteger d = new BigInteger(rd.nextLine());
return d;
}
}
F 題: Non-overlapping Squares
//ANMHLIJKTJIY!
#pragma GCC optimize(2)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#include <bits/stdc++.h>
#define INF 1000000000
#define LINF 1000000000000000000
#define MOD 1000000007
#define mod 998244353
#define F first
#define S second
#define ll long long
#define N 1010
using namespace std;
ll n,m,a[N][N],v[N][N],mx[4][N][N],mxc[N],mxr[N];
ll ask(ll lx,ll ly,ll rx,ll ry)
{
return a[rx][ry]-a[lx-1][ry]-a[rx][ly-1]+a[lx-1][ly-1];
}
int main(){
ll i,j;
cin>>n>>m;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
cin>>a[i][j];
a[i][j]+=a[i][j-1];
}
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
a[i][j]+=a[i-1][j];
}
}
for(i=1;i+m-1<=n;i++)
{
for(j=1;j+m-1<=n;j++)
{
v[i][j]=ask(i,j,i+m-1,j+m-1);
mx[0][i][j]=mx[1][i][j]=mx[2][i][j]=mx[3][i][j]=v[i][j];
mxr[i]=max(mxr[i],v[i][j]);
mxc[j]=max(mxc[j],v[i][j]);
}
}
for(i=1;i<=n-m+1;i++)
{
for(j=1;j<=n-m+1;j++)
{
mx[0][i][j]=max(max(mx[0][i][j],mx[0][i-1][j]),mx[0][i][j-1]);
}
}
for(i=1;i<=n-m+1;i++)
{
for(j=n-m+1;j>0;j--)
{
mx[1][i][j]=max(max(mx[1][i][j],mx[1][i-1][j]),mx[1][i][j+1]);
}
}
for(i=n-m+1;i>0;i--)
{
for(j=1;j<=n-m+1;j++)
{
mx[2][i][j]=max(max(mx[2][i][j],mx[2][i+1][j]),mx[2][i][j-1]);
}
}
for(i=n-m+1;i>0;i--)
{
for(j=n-m+1;j>0;j--)
{
mx[3][i][j]=max(max(mx[3][i][j],mx[3][i+1][j]),mx[3][i][j+1]);
}
}
ll ans=0;
for(i=m+1;i+m+m-1<=n;i++)
{
ll cur=0;
for(j=i+m;j+m-1<=n;j++)
{
cur=max(cur,mxc[j-m]);
ans=max(ans,cur+mx[0][n-m+1][i-m]+mx[1][n-m+1][j]);
}
}
for(i=m+1;i+m+m-1<=n;i++)
{
ll cur=0;
for(j=i+m;j+m-1<=n;j++)
{
cur=max(cur,mxr[j-m]);
ans=max(ans,cur+mx[0][i-m][n-m+1]+mx[2][j][n-m+1]);
}
}
for(i=m;i+m-1<=n;i++)
{
for(j=m;j+m-1<=n;j++)
{
ans=max(ans,mx[0][i-m][j-m]+mx[2][i][j-m]+mx[3][1][j]);
ans=max(ans,mx[2][1][j-m]+mx[1][i-m][j]+mx[3][i][j]);
ans=max(ans,mx[0][i-m][j-m]+mx[1][i-m][j]+mx[3][i][1]);
ans=max(ans,mx[1][i-m][1]+mx[2][i][j-m]+mx[3][i][j]);
}
}
cout<<ans<<'\n';
return 0;
}