性质: 相邻两数互质

$$x=(m+1)/2$$

m = int(input())print ((m + 1) // 2)

性质: 各个位数之和是3的倍数，可被3整除

和数的组合顺序无关

n = int(input())
arr = list(map(int, input().split()))res = 0
for v in arr:while v > 0:res += v % 10v //= 10if res %3 == 0:print ("YES")
else:print ("NO")

思路: 分类讨论题

要注意的是，可以通过耗电回撤到超级快充的状态

x, y, t, a, b, c = list(map(int, input().split()))if x <= t:r = min((100 - x) / c, (100 - x) / a, (100 - x) / b)print ("%.8f" % (r))
else:r = min((100 - x) / a, (100 - x) / b, (x - t) / y + (100 - t) / c)print ("%.8f" % (r)) 

因为a是一个大数

所以从gcd的推演公式切入

$$gcd(a,b)=gcd(b,a$$

可以迭代线性遍历a，来求解a%b的值

然后在求gcd(a%b, b)

a = input()
b = int(input())r = 0
for c in a:p = ord(c) - ord('0')r = (r * 10 + p) % bfrom math import gcdprint (gcd(r, b))

方法1: 并查集

方法2: 二分+bfs

方法3: dijkstra

这边给了并查集的思路，即按权值从小到大合并，直至(0,0)和(n-1,m-1)连通.

class Dsu(object):def __init__(self, n) -> None:self.n = nself.arr = [-1] * ndef find(self, u):if self.arr[u] == -1:return uself.arr[u] = self.find(self.arr[u])return self.arr[u]def merge(self, u, v):a, b = self.find(u), self.find(v)if a != b:self.arr[a] = bn = int(input())vis = [[False] * n for _ in range(n)]
ops = []
grid = []
for i in range(n):row = list(map(int, input().split()))grid.append(row)for j in range(n):ops.append((row[j], i, j))ops.sort(key=lambda x: x[0])ans = 0
dsu = Dsu(n * n)
for (v, y, x) in ops:vis[y][x] = Truefor (dy, dx) in [(1, 0), (-1, 0), (0, 1), (0, -1)]:ty, tx = y + dy, x + dxif 0 <= ty < n and 0 <= tx < n:if vis[ty][tx]:dsu.merge(y * n + x, ty * n + tx)if dsu.find(0) == dsu.find(n * n - 1):ans = vbreakprint (ans)

思路: ST表

这题思路应该蛮多的

有在线解法，离线解法

最直观的

$$区间子数组的绝对值最大和，是区间前缀和的最大值-区间前缀和的最小值$$

• 前缀和
• RMQ(区间最大/最小)

莫队板子也可以的

import java.io.*;
import java.util.StringTokenizer;
import java.util.function.BiFunction;public class Main {public static void main(String[] args) {AReader sc = new AReader();PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));int n = sc.nextInt();int[] arr = new int[n];for (int i = 0; i < n; i++) arr[i] = sc.nextInt();long[] pre = new long[n + 1];for (int i = 0; i < n; i++) pre[i + 1] = pre[i]+arr[i];SparesTable maxSt = new SparesTable(pre, Math::max);SparesTable minSt = new SparesTable(pre, Math::min);int q = sc.nextInt();for (int i = 0; i < q; i++) {int l = sc.nextInt() - 1, r = sc.nextInt();long res = maxSt.query(l, r) - minSt.query(l, r);pw.println(res);}pw.flush();}staticclass SparesTable {long[][] tables;BiFunction<Long, Long, Long> callback;public SparesTable(long[] arr, BiFunction<Long, Long, Long> callback) {int n = arr.length;int m = (int)(Math.log(n) / Math.log(2) + 1);tables = new long[m][n];this.callback = callback;for (int i = 0; i < n; i++) {tables[0][i] = arr[i];}for (int i = 1; i < m; i++) {int half = 1 << (i - 1);for (int j = 0; j + half < n; j++) {tables[i][j] = callback.apply(tables[i - 1][j], tables[i - 1][j + half]);}}}// 闭闭区间long query(int l, int r) {int t = (int)(Math.log(r - l + 1) / Math.log(2));return callback.apply(tables[t][l], tables[t][r - (1 << t) + 1]);}}staticclass AReader {private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));private StringTokenizer tokenizer = new StringTokenizer("");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());}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;}}}