前言
整体评价
C题这个构造题挺好的，赛中把-1写成No, 直接整不会了，T_T.
D题是一道很裸的最小生成树题，只需要一个小小的逆向思维，把删除操作转换为构建过程。
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A. 小红的漂亮串
数据规模较小，直接暴力匹配即可，当然也可以使用API。
import java.io.BufferedInputStream;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
String s = sc.next();
int cnt = 0;
int pos = 0;
while (cnt < 2 && (pos = s.indexOf("red", pos)) != -1) {
pos += 3;
cnt++;
}
System.out.println(cnt >= 2 ? "Yes": "No");
}
}
s = input()
p = 0
cnt = 0
# find 和 index的区别
while cnt < 2 and s.find("red", p) != -1:
cnt += 1
p = s.find("red", p) + 1
print ("Yes" if cnt >= 2 else "No")
B. 小红的偶子串
好像滑窗贪心是错误的
引入dp[i + 1], 表示以i结尾，且是偶数位的最长子串(为了处理方便，漂移了一位)
-
dp[i + 1] = dp[i - 1] + 2, str[i] == str[i - 1]
-
dp[i + 1] = 0, str[i] != str[i - 1]
import java.io.BufferedInputStream;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
char[] str = sc.next().toCharArray();
int n = str.length;
int res = 0;
int[] dp = new int[n + 1];
for (int i = 1; i < n; i++) {
if (str[i] == str[i - 1]) {
dp[i + 1] = dp[i - 1] + 2;
}
res = Math.max(res, dp[i + 1]);
}
System.out.println(res);
}
}
s = input()
n = len(s)
dp = [0] * (n + 1)
dp[0] = 0
for i in range(0, n - 1):
if s[i] == s[i + 1]:
dp[i + 2] = max(dp[i + 2], dp[i] + 2)
print (max(dp))
C. 小红的数组构造
首先可以排除掉不可能的情况
- 1~n的累加和 > x
- n > k
然后想如何构造这个神奇的数组呢？
可以先安排，1~n填充，即arr[i] = i
那这样还剩余
left = x - (n+1)*n/2
那如何处理这个剩余的部分呢？
其实可以对n进行乘除
- d = left / n
- r = left % n
把d赋值给每个元素，然后r这个尾巴相当于给最后的r个元素额外+1
这样的构造数组，应该是理想上最满足要求的数组了。
题外话：
赛中也采用了，类似后悔堆的思路，即先填充1~n，然后多余的从大到小进行置换，可以过，但是没法证明。
import java.io.BufferedInputStream;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
int n = sc.nextInt();
long k = sc.nextLong(), x = sc.nextLong();
long s = ((long)n) * (n + 1) / 2;
if (s > x || n > k) {
System.out.println(-1);
} else {
long[] arr = new long[n + 1];
for (int i = 1; i <= n; i++) {
arr[i] = i;
}
x -= s;
long d = x / n;
long v = x % n;
for (int i = 1; i <= n; i++) {
arr[i] += d;
if (n - i < v) {
arr[i]++;
}
}
// 对数组进行最后的check，保证每个元素都小于等于k
boolean ok = true;
for (int i = 1;i <= n; i++) {
if (arr[i] > k) ok = false;
}
if (!ok) System.out.println("-1");
else System.out.println(IntStream.range(1, n + 1).mapToObj(t -> String.valueOf(arr[t])).collect(Collectors.joining(" ")));
}
}
}
D. 小红的图上删边
逆向思维，把删除转换为重头构建树，那这样就变成求最小生成树的过程了。
这边采用 kruskal算法，借助并查集来实现最小生成树的构建。
这边需要额外计算边权，其实就是统计每个点的2,5的因子数
w(u, v) = min(cnt5(u) + cnt5(v), cnt2(u) + cnt2(v))
整体的时间复杂度为, 主要在排序这块
import java.io.BufferedInputStream;
import java.util.*;
public class Main {
static class Dsu {
int n;
int[] arr;
int[] gz;
public Dsu(int n) {
this.n = n;
this.arr = new int[n + 1];
this.gz = new int[n + 1];
Arrays.fill(gz, 1);
}
int find(int u) {
if (arr[u] == 0) return u;
return arr[u] = find(arr[u]);
}
void union(int u, int v) {
int ai = find(u), bi = find(v);
if (ai != bi) {
arr[ai] = bi;
gz[bi] += gz[ai];
}
}
int gSize(int u) {
return gz[find(u)];
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
int n = sc.nextInt(), m = sc.nextInt();
Dsu dsu = new Dsu(n);
long[] ws = new long[n + 1];
int[] cnt5 = new int[n + 1];
int[] cnt2 = new int[n + 1];
for (int i = 1; i <= n; i++) {
ws[i] = sc.nextLong();
long v = ws[i];
while (v % 5 == 0) {
cnt5[i]++;
v /= 5;
}
while (v % 2 == 0) {
cnt2[i]++;
v /= 2;
}
}
long res = 0;
List<int[]> edges = new ArrayList<>();
for (int i = 0; i < m; i++) {
int u = sc.nextInt(), v = sc.nextInt();
int w = Math.min(cnt5[u] + cnt5[v], cnt2[u] + cnt2[v]);
// 引入边权
edges.add(new int[] {u, v, w});
res += w;
}
// 按边权从小到大排序
Collections.sort(edges, Comparator.comparing(x -> x[2]));
long tmp = 0;
for (int[] e: edges) {
int u = e[0], v = e[1], w = e[2];
if (dsu.find(u) != dsu.find(v)) {
dsu.union(u, v);
tmp += w;
if (dsu.gSize(1) == n) {
break;
}
}
}
// 删边收益 = 总的权值 - 最小生成树的权值
System.out.println(res - tmp);
}
}
import heapq
class Dsu(object):
def __init__(self, n: int):
self.n = n
self.arr = [0] * (n + 1) # index-1
self.group = n
def find(self, u: int) -> int:
if self.arr[u] == 0:
return u
self.arr[u] = self.find(self.arr[u])
return self.arr[u]
def union(self, u :int, v: int):
a = self.find(u)
b = self.find(v)
if a != b:
self.arr[a] = b
self.group -= 1
def groupNum(self) -> int:
return self.group
n, m = list(map(int, input().split()))
arr = list(map(int, input().split()))
# 逆向思维
cnt2 = [0] * (n + 1)
cnt5 = [0] * (n + 1)
for i in range(len(arr)):
v = arr[i]
while v % 2 == 0:
v /= 2
cnt2[i + 1] += 1
while v % 5 == 0:
v /= 5
cnt5[i + 1] += 1
res = 0
edges = []
for i in range(m):
u, v = list(map(int, input().split()))
edges.append((min(cnt2[u] + cnt2[v], cnt5[u] + cnt5[v]), u, v))
res += min(cnt2[u] + cnt2[v], cnt5[u] + cnt5[v])
dsu = Dsu(n)
heapq.heapify(edges)
while dsu.groupNum() > 1 and len(edges):
row = heapq.heappop(edges)
if dsu.find(row[1]) != dsu.find(row[2]):
dsu.union(row[1], row[2])
res -= row[0]
print (res)
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