1. Perfect number
4876. Perfect Number - AcWing Question Bank
1. Topic
A positive integer is said to be perfect if it is divisible by 2520.
Given a positive integer n, please calculate how many perfect numbers there are in the range [1,n].
input format
an integer n.
output format
An integer representing the number of perfect numbers in the range [1,n].
data range
The first 3 test points satisfy 1≤n≤3000.
All test points satisfy 1≤n≤10¹⁸.
Input sample:
3000
Sample output:
12. Ideas
Simple questions, just look at the code
3. Code
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
long n=sc.nextLong();
System.out.println(n/2520);
}
}
2. Maximum value
4877. Maximum Value - AcWing Question Bank
1. Topic
data range
The first 4 test points satisfy 1≤n≤100, 1≤m≤2.
All test points satisfy 1≤n≤1000, 1≤m≤10, 1≤ lᵢ ,hᵢ ,vᵢ ,wᵢ ≤100.
Input sample 1:
10 2 2 1
7 3 2 100
12 3 1 10
Output sample 1:
241
Input sample 2:
100 1 25 50
15 5 20 10
Output sample 2:
2002. Ideas
If you see a particularly well-written solution in the solution, let's just read it!
Backpack Problem - 01 Backpack|Complete Backpack_Leng Xixue's Blog-CSDN Blog
You can take a look at this 01 backpack for a better understanding.
3. Code
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int v0 = sc.nextInt();
int w0 = sc.nextInt();
long[] count = new long[n + 1];//背包的容量
for (int i = 1; i <= m; i++) {//先遍历第1到m个物品
int l = sc.nextInt(), h = sc.nextInt(), v = sc.nextInt(), w = sc.nextInt();
int s = l / h;//计算物品i可以放几个
for (int j = n; j >= 0; j--) {//再倒序遍历背包容量
for (int k = 1; k <= s && k * v <= j; k++) {//1到s个物品i放与不放 取最大值
count[j] = Math.max(count[j], count[j - k * v] + (long) k * w);
}
}
}
long ans = 0;
for (int i = 0; i <= n; i++) {//遍历背包容量，先看是否可以放进，再取放与不放的最大值
if (i >= v0) {
count[i] = Math.max(count[i], count[i - v0] + w0);
}
//求出最大的背包价值
ans = Math.max(ans, count[i]);
}
System.out.println(ans);
}
}3. Maintain the array
4878. Maintain Array - AcWing Question Bank
1. Topic
input format
The first line contains 55 integers n,k,a,b,q.
The next q lines, each line describes an operation, the format is as described in the title.
output format
For each operation, output a line with an integer representing the result. 2 p
data range
The first 33 test points satisfy 1≤k≤n≤10, 1≤q≤10.
All test points satisfy 1≤k≤n≤2×10⁵, 1≤b<a≤10000, 1≤q≤2×10⁵, 1≤x≤n, 1≤y≤10000, 1≤p≤n−k+ 1.
Input sample 1:
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output sample 1:
3
6
4
Input sample 2:
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output sample 2:
7
12. Ideas
To make the time complexity of the written code logN, we need to use a tree array. Ordinary methods will run with a timeout.
If you want to know more about tree arrays, you can read this blog
Simple and easy-to-understand detailed explanation of tree arrays - FlushHip's Blog - CSDN Blog
If you want to quickly understand the idea of writing questions, you can read this blog
Detailed Explanation of Tree Array - Xiao Heshan's Blog - CSDN Blog
3. Code
import java.io.*;
public class Main {
static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
static int N = 200010, n, k, a, b, q;
static int[] tree1 = new int[N], tree2 = new int[N], d = new int[N];
static int lowbit(int x) {
return x & -x;
}
static void add(int[] tree, int x, int v) {
for (int i = x; i <= n; i += lowbit(i))
tree[i] += v;
}
static long query(int[] tree, int x) {
long ans = 0;
for (int i = x; i > 0; i -= lowbit(i))
ans += tree[i];
return ans;
}
public static void main(String[] args) throws Exception{
String[] s1 = in.readLine().split(" ");
n = Integer.parseInt(s1[0]);
k = Integer.parseInt(s1[1]);
a = Integer.parseInt(s1[2]);
b = Integer.parseInt(s1[3]);
q = Integer.parseInt(s1[4]);
while (q -- > 0) {
String[] s2 = in.readLine().split(" ");
int op = Integer.parseInt(s2[0]);
if (op == 1) { //增加 进行单点修改
int x = Integer.parseInt(s2[1]), y = Integer.parseInt(s2[2]);
add(tree1, x, Math.min(d[x] + y, b) - Math.min(d[x], b));
add(tree2, x, Math.min(d[x] + y, a) - Math.min(d[x], a));
d[x] += y;
}else { //查询 进行区间查询
int p = Integer.parseInt(s2[1]);
long sum = query(tree1, p - 1) + query(tree2, n) - query(tree2, p + k - 1);
out.println(sum);
}
}
out.flush();
}
}