目录
题目地址
牛客竞赛_ACM/NOI/CSP/CCPC/ICPC算法编程高难度练习赛_牛客竞赛OJ
做题情况
A 题
输出两个不是同一方位的字符中的任意一个就行
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static IOS sc=new IOS();
static final int MOD = 998244353;
public static void solve() throws IOException {
char a=sc.nextChar();
if(a=='U'||a=='D') {
dduoln('L');
}else {
dduoln('U');
}
}
public static void main(String[] args) throws Exception {
int t = 1;
// t = sc.nextInt();
while (t-- > 0) {solve();}
}
static <T> void dduo(T t) {System.out.print(t);}
static <T> void dduoln(T t) {System.out.println(t);}
}
class IOS{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public IOS(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public BigDecimal nextDecimal() throws IOException{
return new BigDecimal(next());
}
}B 题
可以稍微贪心一下
构造 1 2 1 2 1 2 1 2 1 2
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static IOS sc=new IOS();
static final int MOD = 998244353;
public static void solve() throws IOException {
int n=sc.nextInt();
for(int i=0;i<n;i++) {
if(i%2==0) {
dduo(1+" ");
}else {
dduo(2+" ");
}
}
dduoln("");
}
public static void main(String[] args) throws Exception {
int t = 1;
t = sc.nextInt();
while (t-- > 0) {solve();}
}
static <T> void dduo(T t) {System.out.print(t);}
static <T> void dduoln(T t) {System.out.println(t);}
}
class IOS{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public IOS(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public BigDecimal nextDecimal() throws IOException{
return new BigDecimal(next());
}
}C 题
通过打表发现
1212/12
123123/123
112112/112
都能除尽
所以 我 直接 (str+""+str) /str 求出结果
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static IOS sc=new IOS();
static final int MOD = 998244353;
public static void solve() throws IOException {
String str=sc.next();
BigInteger b=new BigInteger(str+str);
BigInteger c=new BigInteger(str);
dduoln(b.divide(c));
}
public static void main(String[] args) throws Exception {
int t = 1;
t = sc.nextInt();
while (t-- > 0) {solve();}
}
static <T> void dduo(T t) {System.out.print(t);}
static <T> void dduoln(T t) {System.out.println(t);}
}
class IOS{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public IOS(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public BigDecimal nextDecimal() throws IOException{
return new BigDecimal(next());
}
}D 题
模拟题
我的解法是先求出位于 (1,1)(2,2)(3,3)(4,4) 点要多少步
不难是 9 ，25， 49， 81...步
我们发现是奇数的平方
所以先求出要求的这个点是从哪个开始的 即(a，a)
然后算到(a,a)一共要多少步
求出还剩多少步
剩下了多少步然后是 在上面走 在右边走 在下面走 在左边走 在上面走 根据 a 可以算出每条边各需要多少步
在每条边上更新剩下要走的步数
即可求出
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static IOS sc=new IOS();
static final int MOD = 998244353;
public static void solve() throws IOException {
long n=sc.nextLong();
long ii=0;
ii=(long) Math.sqrt(n);
ii+=1;
if(ii%2==0) {
ii++;
}
ii-=2;
// dduoln(ii);
//
// for(int i=1;i<=1e9;i+=2) {
// if(i*i>n) {
// ii=i-2;
// break;
// }
// }
// dduoln(ii);
long a=(ii-1)/2; // 起始位置(a,a)
// dduoln(a+" "+a);
long ans=(n-ii*ii); // 剩下多少步
// dduoln(ans);
if(ans==0) {
dduoln(a+" "+a);
return;
}else if(ans==1) {
dduoln( (a+1) +" "+a);
return;
}
long duanbian=a*2+1;
long changbian=(a+1)*2;
ans-=1;
// 右
if(ans<duanbian) {
dduoln((a+1)+" "+(a-ans));
return;
}
ans-=duanbian;
// 下
if(ans<changbian) {
dduoln((a+1-ans)+" "+(-a-1));
return;
}
ans-=changbian;
// 左
if(ans<changbian) {
dduoln((-a-1)+" "+(-a-1+ans));
return;
}
ans-=changbian;
dduoln((-a-1+ans)+" "+(a+1));
}
public static void main(String[] args) throws Exception {
int t = 1;
t = sc.nextInt();
while (t-- > 0) {solve();}
}
}
class IOS{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public IOS(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public BigDecimal nextDecimal() throws IOException{
return new BigDecimal(next());
}
}E 题
动态规划
首先知道了 第一次可以到达 1 6
第二次可以到达 2 12
第三次可以到达 3 18
以此类推
维护一个 dp 数组 代表的是到达当前位置获取到的最大贡献值
再在每次运动 都要 for(1-6) 找一下这六步对 dp 数组的更新
需要注意的是 不要访问到数组外面去
最后找出 dp 数组中的最大值就行
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static IOS sc=new IOS();
static final int MOD = 998244353;
public static void solve() throws IOException {
int n = sc.nextInt(); // n个数
int k = sc.nextInt(); // 两次
int[] a = new int[n]; // 每个数的数值
for (int i = 0; i < n; i++) {
a[i] = sc.nextInt();
}
long[] dp = new long[n + 1];
Arrays.fill(dp, Long.MIN_VALUE);
dp[0] = 0;
for (int step = 1; step <= k; step++) {
long[] current = new long[n + 1];
Arrays.fill(current, Long.MIN_VALUE);
// 求出当前可到达的位置
int xmin = step - 1;
int xmax = Math.min(6 * (step - 1), n);
// dduoln(xmin+" "+xmax);
for (int x = xmin; x <= xmax; x++) {
for (int d = 1; d <= 6; d++) {
int j = x + d;
if (j > n) {
continue;
}
current[j]=Math.max(current[j],dp[x] + a[j - 1]);
}
}
dp = current;
}
int maxfoot = Math.min(6 * k, n); // 最大步数
long max = Long.MIN_VALUE;
for (int i = k; i <= maxfoot; i++) {
max=Math.max(dp[i], max);
}
dduoln(max);
}
public static void main(String[] args) throws Exception {
int t = 1;
// t = sc.nextInt();
while (t-- > 0) {solve();}
}
static <T> void dduo(T t) {System.out.print(t);}
static <T> void dduoln(T t) {System.out.println(t);}
}
class IOS{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public IOS(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public BigDecimal nextDecimal() throws IOException{
return new BigDecimal(next());
}
}F 题
BFS
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static IOS sc=new IOS();
static final int MOD = 998244353;
public static void solve() throws IOException {
int startX=0, startY=0, endX=0, endY=0;
int n = sc.nextInt();
int m = sc.nextInt();
int h = sc.nextInt();
char[][] grid = new char[n + 1][m + 1];
int[][] minTime = new int[n + 1][m + 1];
for (int i = 0; i <= n; i++) {
Arrays.fill(minTime[i], Integer.MAX_VALUE);
}
for (int i = 1; i <= n; i++) {
String s = sc.next();
for (int j = 1; j <= m; j++) {
grid[i][j] = s.charAt(j-1);
if (grid[i][j] == '*') { // 记录水源的位置
startX = i;
startY = j;
} else if (grid[i][j] == '%') { // 记录终点的位置
endX = i;
endY = j;
}
}
}
int ans=0;
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
pq.offer(new int[]{0, startX, startY, 1});
while (!pq.isEmpty()) {
int[] cur = pq.poll();
int time = cur[0], x = cur[1], y = cur[2], state = cur[3];
if (x == endX && y == endY) { // 到达了终点
ans=time;
break;
}
if (state == 0 && minTime[x][y] <= time) continue;
minTime[x][y] = Math.min(minTime[x][y], time);
if (x + 1 <= n) {
if (grid[x + 1][y] == '#') {
if (y - 1 >= 1 && grid[x][y - 1] != '#') {
pq.offer(new int[]{time + 1, x, y - 1, 0});
}
if (y + 1 <= m && grid[x][y + 1] != '#') {
pq.offer(new int[]{time + 1, x, y + 1, 0});
}
if (state == 1) {
pq.offer(new int[]{time + h + 1, x + 1, y, 1});
}
}
else {
pq.offer(new int[]{time + 1, x + 1, y, 1});
}
}
ans=Integer.MAX_VALUE;
}
dduoln(ans == Integer.MAX_VALUE ? -1 : ans);
}
public static void main(String[] args) throws Exception {
int t = 1;
// t = sc.nextInt();
while (t-- > 0) {solve();}
}
static <T> void dduo(T t) {System.out.print(t);}
static <T> void dduoln(T t) {System.out.println(t);}
}
class Node implements Comparable<Node> {
int x, y, nowT, abi;
public Node(int x, int y, int nowT, int abi) {
this.x = x;
this.y = y;
this.nowT = nowT;
this.abi = abi;
}
@Override
public int compareTo(Node other) {
return Integer.compare(other.nowT, this.nowT); // 按照 nowT 值进行比较，形成最大堆
}
}
class IOS{
BufferedReader bf;
StringTokenizer st;
BufferedWriter bw;
public IOS(){
bf=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
bw=new BufferedWriter(new OutputStreamWriter(System.out));
}
public String nextLine() throws IOException{
return bf.readLine();
}
public String next() throws IOException{
while(!st.hasMoreTokens()){
st=new StringTokenizer(bf.readLine());
}
return st.nextToken();
}
public char nextChar() throws IOException{
return next().charAt(0);
}
public int nextInt() throws IOException{
return Integer.parseInt(next());
}
public long nextLong() throws IOException{
return Long.parseLong(next());
}
public double nextDouble() throws IOException{
return Double.parseDouble(next());
}
public float nextFloat() throws IOException{
return Float.parseFloat(next());
}
public BigInteger nextBigInteger() throws IOException{
return new BigInteger(next());
}
public BigDecimal nextDecimal() throws IOException{
return new BigDecimal(next());
}
}