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裁纸刀

原题链接

问题描述

解题思路

找规律，首先需要裁掉纸的边缘 4 刀，再看横向的，要裁 行数 - 1 刀，纵向的每一行都要裁掉 列数 - 1 刀。

参考代码

import java.io.*;import java.util.*;public class Main {
       static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));        public static void main(String[] args) throws Exception {
       	    	out.println(4 + (20 - 1) + (21 * 20));        out.flush();        in.close();    }}

刷题统计

原题链接

问题描述

解题思路

数据范围很大，需要把一个周的刷题量看作一个整体来计算。

参考代码

import java.io.*;import java.util.*;public class Main {
       static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));        public static void main(String[] args) throws Exception {
       		    	String[] s = in.readLine().split(" ");    	long a = Long.parseLong(s[0]), b = Long.parseLong(s[1]), p = Long.parseLong(s[2]);    	// 一周的刷题数量    	long weekCnt = a * 5 + b * 2;    	// 刷题的周数    	long weekSum = p / weekCnt;    	// 当前刷题天数    	long ans = weekSum * 7;    	// 剩余的题目    	p %= weekCnt;    	// 当前是星期一    	int cur = 1;    	while (p > 0) {
       		p -= cur <= 5? a: b;    		cur++;    		ans++;    	}    	out.println(ans);        out.flush();        in.close();    }}

修剪灌木

原题链接

问题描述

解题思路

对于当前第 i 棵树，再修剪完之后下一次修剪的时间从两边取一个最大值即可，可以发现是左右对称的，计算左边再对称右边即可，省去了判断大小。

参考代码

import java.io.*;public class Main {
       static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));        public static void main(String[] args) throws Exception {
       		    	int n = Integer.parseInt(in.readLine());    	int mid = n / 2;    	for (int i = 1; i <= mid; i++) {
       		out.println((n - i) * 2);    	}    	    	if ((n & 1) == 1) out.println((n - mid - 1) * 2);    	    	for (int i = mid; i >= 1; i--) {
       		out.println((n - i) * 2);    	}        out.flush();        in.close();    }}

k倍区间

原题链接

问题描述

解题思路

前缀和 + 哈希表

参考代码

import java.io.*;import java.util.*;public class Main {
       static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));        static int N = (int)1e5 + 10, n, k;    static int[] a = new int[N], pre = new int[N];    static Map
   
     map = new HashMap<>();        public static void main(String[] args) throws Exception {
       	String[] s = in.readLine().split(" ");    	n = Integer.parseInt(s[0]);    	k = Integer.parseInt(s[1]);    	for (int i = 1; i <= n; i++) {
       		a[i] = Integer.parseInt(in.readLine());    		pre[i] = (pre[i - 1] + a[i]) % k;    	}    	long ans = 0L;    	for (int i = 0; i <= n; i++) {
       		ans += map.getOrDefault(pre[i], 0);    		map.put(pre[i], map.getOrDefault(pre[i], 0) + 1);    	}    	    	out.println(ans); 	        out.flush();        in.close();    }}
   

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