1.约数个数
由乘法原理可以得出：
import java.util.*;
public class Main{
static int mod = (int)1e9 + 7;
public static void main(String[] args){
Map<Integer,Integer> map = new HashMap<>(); //创建一个哈希表
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
while(n -- > 0){
int x = scan.nextInt();
//下面这里是运用了分解质因数的模板，
for(int i = 2 ; i <= x / i ; i ++ ){
while(x % i == 0){
x /= i;
// map.getOrDefault(i,0) 这个是获取对应i位置的values值
map.put(i,map.getOrDefault(i,0) + 1);
}
}
if(x > 1) map.put(x,map.getOrDefault(x,0) + 1 );
}
long res = 1;
//map.keySet()获取所有的key值,map.values()获取所有的values值，两种方法都可以
for(int key : map.values()){
res = res * (key + 1) % mod;
}
System.out.println(res);
}
}2.堆优化版的Dijkstra
正好复习一下优先队列和存图方式;
3.贡献法求因数个数和；
题目来源于华北水利水电大学校赛：
7-10 兔丁兴旺的兔子家族 - 2024年第六届华北水利水电大学校赛-正式赛-复盘 (pintia.cn)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
long sum=0;
for(int i=1;i<=n;i++) {
sum+=n/i;
}
System.out.println(sum);
}
}4.BFS--青蛙跳杯子
开心开心，一次就ac了，嘿嘿；
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class Main {
static String start, end;
static int[] dx = {-3,-2,-1,1,2,3 };
static int n;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
start = sc.next();
end = sc.next();
n = start.length();
bfs();
}
public static void bfs() {
Queue<String>q=new LinkedList<String>();
q.offer(start);
HashMap<String, Integer>map=new HashMap<String, Integer>();
map.put(start, 0);
while(!q.isEmpty()) {
String t=q.poll();
if(t.equals(end)) {
System.out.println(map.get(t));
}
int index=t.indexOf('*');
for(int i=0;i<6;i++) {
int x=index+dx[i];
if(x<0||x>=n)continue;
char[] T = t.toCharArray();
char tmp=T[x];
T[x]=T[index];
T[index]=tmp;
String str=new String(T);
if(!map.containsKey(str)) {
map.put(str, map.get(t)+1);
}else continue;
q.offer(str);
}
}
}
}5.进制转换
import java.util.Scanner;
public class Excel地址 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int []a=new int[26];
int i=0;
while(n!=0) {
n--;
a[i++]=n%26;
n/=26;
}
for(int j=i-1;j>=0;j--) {
System.out.print((char)(a[j]+65));
}
}
}6.二分答案---
import java.util.ArrayList;
import java.util.Scanner;
public class Main {
static int N = 100010;
static long[] a = new long[N];
static long[] b = new long[N];
static int n;
static int k;
static int INF = Integer.MAX_VALUE;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
k = sc.nextInt();
for (int i = 1; i <= n; i++) {
a[i] = sc.nextLong();
}
for (int i = 1; i <= n; i++) {
b[i] = sc.nextLong();
}
long l = 0, r =(int)2e9;
while (l < r) {
long mid = (l + r + 1) / 2;
if (check(mid)) {
l = mid;
} else
r = mid - 1;
}
System.out.println(l); // 二分逻辑
}
public static boolean check(long x) {
long m=k;
for (int i = 1; i <= n; i++) {
if (b[i] - x * a[i] < 0) {
m-=(a[i]*x-b[i]);
if(m<0)return false;
}
}
return true;
}
}7.FloodFill
import java.util.Scanner;
public class Main {
static int n,m;
static int N=110;
static int[][]g=new int[N][N];
static boolean[][]st=new boolean[N][N];
static int res=0;
static int[] dx = { -1, 0, 1, 0 };
static int[] dy = { 0, 1, 0, -1 };
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
m=sc.nextInt();
for(int i=0;i<n;i++) {
String str = sc.next();
for(int j=0;j<m;j++) {
g[i][j]=str.charAt(j);
}
}
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
if(g[i][j]=='B'&&!st[i][j]) {
res++;
dfs(i,j);
}
}
}
System.out.println(res);
}
public static void dfs(int a,int b) {
st[a][b]=true;
for(int i=0;i<4;i++) {
int x=a+dx[i];
int y=b+dy[i];
if(x<0||x>=n||y<0||y>=m||st[x][y]||g[x][y]=='.')
continue;
dfs(x,y);
}
}
}