前言
整体评价
F题真心好题，很典，学到了很多。D题用了对顶堆，写到一半就想到了更简单的方法，哭。E题是基于众数的构造。
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A. 小红大战小紫
思路: 模拟
n, m = list(map(int, input().split()))
if n > m:
print("kou")
elif n < m:
print("yukari")
else:
print("draw")
B. 小红的白日梦
思路: 模拟
n = int(input())
day = input()
night = input()
res = 0
for i in range(n):
if day[i] == 'Y' and night[i] == 'Y':
res += 3
elif day[i] == 'Y' and night[i] == 'N':
res += 2
elif day[i] == 'N' and night[i] == 'Y':
res += 2
print(res)
C. 小红的小小红
思路: 贪心构造
先构造xiaohong, 后续字符按剩余数量追加即可
from collections import Counter
s = input()
cnt = Counter(s)
res = []
for c in "xiaohong":
res.append(c)
cnt[c] -= 1
for (k, v) in cnt.items():
if v > 0:
res.extend([k] * v)
print(''.join(res))
D. 小红的中位数
思路: 对顶堆
写复杂了，中间状态维护太麻烦了。
仅供负面教材典型。
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.stream.Collectors;
public class Main {
public static void main(String[] args) {
AReader sc = new AReader();
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
// 对顶堆
TreeMap<Integer, Integer> h1 = new TreeMap<>();
TreeMap<Integer, Integer> h2 = new TreeMap<>();
int cnt1 = 0, cnt2 = 0;
for (int i = 0; i < n; i++) {
if (cnt1 == cnt2) {
h1.merge(arr[i], 1, Integer::sum);
cnt1++;
} else {
h2.merge(arr[i], 1, Integer::sum);
cnt2++;
}
if (cnt1 > 0 && cnt2 > 0 && h1.lastKey() > h2.firstKey()) {
int k1 = h1.lastKey();
int k2 = h2.firstKey();
h1.computeIfPresent(k1, (a, b) -> b > 1 ? b - 1 : null);
h2.computeIfPresent(k2, (a, b) -> b > 1 ? b - 1 : null);
h1.merge(k2, 1, Integer::sum);
h2.merge(k1, 1, Integer::sum);
}
}
double[] f = new double[n];
for (int i = 0; i < n; i++) {
if (h1.containsKey(arr[i])) {
h1.computeIfPresent(arr[i], (a, b)-> b > 1 ? b - 1 : null);
cnt1--;
} else {
h2.computeIfPresent(arr[i], (a, b)-> b > 1 ? b - 1 : null);
cnt2--;
}
if (cnt1 < cnt2) {
int k2 = h2.firstKey();
h2.computeIfPresent(k2, (a, b) -> b > 1 ? b - 1 : null);
h1.merge(k2, 1, Integer::sum);
cnt1++;
cnt2--;
} else if (cnt1 > cnt2 + 1) {
int k1 = h1.lastKey();
h1.computeIfPresent(k1, (a, b) -> b > 1 ? b - 1 : null);
h2.merge(k1, 1, Integer::sum);
cnt1--;
cnt2++;
}
if (cnt1 > cnt2) {
f[i] = h1.lastKey();
} else {
f[i] = (h1.lastKey() + h2.firstKey()) / 2.0;
}
if (cnt1 > cnt2) {
h2.merge(arr[i], 1, Integer::sum);
cnt2++;
} else {
h1.merge(arr[i], 1, Integer::sum);
cnt1++;
}
while (cnt1 > 0 && cnt2 > 0 && h1.lastKey() > h2.firstKey()) {
int k1 = h1.lastKey();
int k2 = h2.firstKey();
h1.computeIfPresent(k1, (a, b) -> b > 1 ? b - 1 : null);
h2.computeIfPresent(k2, (a, b) -> b > 1 ? b - 1 : null);
h1.merge(k2, 1, Integer::sum);
h2.merge(k1, 1, Integer::sum);
}
}
System.out.println(Arrays.stream(f).mapToObj(x -> String.format("%.1f", x)).collect(Collectors.joining("\n")));
}
static
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
E. 小红构造数组
思路: 众数+构造
只要 众数 <=(m+1)/2, m为数组长度, 则一定可以构造
按频率(从大到小)优先填充偶数位(0-index)，然后在填充奇数位，这样天然保证相同元素被隔离
可以反证，证明该思路，一定是OK的
import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
public class Main {
static List<long[]> split(long n) {
List<long[]> res = new ArrayList<>();
for (long i = 2; i <= n / i; i++) {
if (n % i == 0) {
int cnt = 0;
while (n % i == 0) {
n /= i;
cnt++;
}
res.add(new long[] {i, cnt});
}
}
if (n > 1) {
res.add(new long[] {n, 1});
}
return res;
}
public static void main(String[] args) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
long e = sc.nextLong();
if (e == 1) {
System.out.println(-1);
return;
}
List<long[]> primes = split(e);
// 感觉像众数的构造
int n = 0;
Collections.sort(primes, Comparator.comparing(x -> -x[1]));
int m = (int)primes.get(0)[1];
for (long[] x: primes) {
n += (int)x[1];
}
if (m > (n + 1) / 2) {
System.out.println(-1);
} else {
Long[] res = new Long[n];
int ptr = 0;
for (long[] x: primes) {
for (int j = 0; j < x[1]; j++) {
res[ptr] = x[0];
ptr += 2;
if (ptr >= n) ptr = 1;
}
}
System.out.println(n);
System.out.println(Arrays.stream(res).map(String::valueOf).collect(Collectors.joining(" ")));
}
}
}
F. 小红又战小紫
背景: 博弈+概率DP
思路: 状态压缩 + 博弈简化
因为和堆的位子无关，只和堆的石头数有关。
因此可以把很多堆(只有1和2)，压缩为(1个石头的堆数，2个石头的堆数)
这样在博弈过程中，其状态就从n维(n<=1000)降维压缩到2维
博弈简化
博弈是套着 alpha+beta剪枝
但是这里面其实有策略挑选
- 如果选用1技能，得到胜率概率为p1
- 如果选用2技能，得到胜率概率为p2
取其 max(p1, p2)
因为概率取模会丢失大小关系，所以这边需要保留真实的概率值，这样就很麻烦。
这边有个特殊情况
就是当1,2个数得石头堆都存在时，如果选用技能2，那一定输，也就是胜率p2=0
在这个前提下
博弈策略，只需要单独评估技能1就行(除只有1个石头堆的情况)，这样就可以runtime过程中，使用模。
- 记忆化搜索
dp[i][j] = j / inv(i + j) * (1 - dp[i+1][j-1]) + i / inv(i+j) * (1 - dp[i - 1][j])
- 题外话
如果这题没有限制石头数目为1~2之间，那么这题得保留真实的概率，而不是模意义下的概率。
java实际测试保留真实概率，大概能过13.3%，然后WA(整数溢出了).
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;
public class Main {
static long mod = (long)1e9 + 7;
static int MZ = 1000;
static long[] gFac = new long[MZ + 1];
static long[] gInv = new long[MZ + 1];
static {
gFac[0] = 1;
for (int i = 1; i <= MZ; i++) {
gFac[i] = gFac[i-1] * i % mod;
}
gInv[MZ] = BigInteger.valueOf(gFac[MZ]).modInverse(BigInteger.valueOf(mod)).longValue();
for (int i = MZ - 1; i >= 0; i--) {
gInv[i] = gInv[i + 1] * (i + 1) % mod;
}
}
static Map<Long, Long> memo = new HashMap<>();
static long dfs(int x, int y) {
// 直接使用技能2
if (x > 0 && y == 0) {
return 1;
}
// 必输态
if (x == 0 && y == 0) {
return 0;
}
long k = ((long)x << 32) | y;
if (memo.containsKey(k)) {
return memo.get(k);
}
long inv = gInv[x + y] * gFac[x + y - 1] % mod;
// 策略1，随机堆
long r = 0;
if (x > 0) {
long p = dfs(x - 1, y);
r += (1 - p) * x % mod * inv % mod;
}
if (y > 0) {
long p = dfs(x + 1, y - 1);
r += (1 - p) * y % mod * inv % mod;
}
r = (r % mod + mod) % mod;
memo.put(k, r);
return r;
}
public static void main(String[] args) {
AReader sc = new AReader();
int x = 0, y = 0;
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
if (arr[i] == 1) x++;
else if (arr[i] == 2) y++;
}
long p = dfs(x, y);
System.out.println(p);
}
static
class AReader {
private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
private StringTokenizer tokenizer = new StringTokenizer("");
private String innerNextLine() {
try {
return reader.readLine();
} catch (IOException ex) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String nextLine = innerNextLine();
if (nextLine == null) {
return false;
}
tokenizer = new StringTokenizer(nextLine);
}
return true;
}
public String nextLine() {
tokenizer = new StringTokenizer("");
return innerNextLine();
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
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