当前位置：首页 > news >正文

做网站需要交维护费么百度推广工作好干吗

news 2025/7/29 1:08:38

做网站需要交维护费么,百度推广工作好干吗,网站维护中要多久才能重新进入,做app和网站哪个D - Square Pair 题目大意给一长为的数组，问有多少对，两者相乘为非负整数完全平方数解题思路一个数除以其能整除的最大的完全平方数，看前面有多少个与其余数相同的数，两者乘积满足条件（已经是完全平方数的部分无…

D - Square Pair

题目大意

给一长为 $n$ 的数组 $a$ ，问有多少对 $A_i,A_j(1\leq i< j\leq n)$ ，两者相乘为非负整数完全平方数

解题思路

一个数除以其能整除的最大的完全平方数，看前面有多少个与其余数相同的数，两者乘积满足条件（已经是完全平方数的部分无用）
对于0，特判（与%=0区分开）

import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Vector;public class Main{static long md=(long)998244353;static long Linf=Long.MAX_VALUE/2;static int inf=Integer.MAX_VALUE/2;public static void main(String[] args) throws IOException{AReader input=new AReader();PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));int n=input.nextInt();int[] a=new int[n+1];HashMap<Integer, Integer> hs=new HashMap<Integer, Integer>();long ans=0;for(int i=1;i<=n;++i) {a[i]=input.nextInt();if(a[i]==0) {continue;}int t=a[i];int y=(int)Math.sqrt(a[i]);while(y>0) {int z=y*y;if(t%z==0) {t/=z;break;}y--;}if(hs.get(t)!=null) {int p=hs.get(t);ans+=p;hs.put(t, p+1);}else {hs.put(t, 1);}}int zero=0;for(int i=1;i<=n;++i) {if(a[i]==0) {ans+=i-1;zero++;}else {ans+=zero;}}out.print(ans);out.flush();out.close();}//System.out.println();//out.println();//String o="abcdefghijklmnopqrstuvwxyz";//char[] op=o.toCharArray();staticclass AReader{ BufferedReader bf;StringTokenizer st;BufferedWriter bw;public AReader(){bf=new BufferedReader(new InputStreamReader(System.in));st=new StringTokenizer("");bw=new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException{return bf.readLine();}public String next() throws IOException{while(!st.hasMoreTokens()){st=new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException{//确定下一个token只有一个字符的时候再用return next().charAt(0);}public int nextInt() throws IOException{return Integer.parseInt(next());}public long nextLong() throws IOException{return Long.parseLong(next());}public double nextDouble() throws IOException{return Double.parseDouble(next());}public float nextFloat() throws IOException{return Float.parseFloat(next());}public byte nextByte() throws IOException{return Byte.parseByte(next());}public short nextShort() throws IOException{return Short.parseShort(next());}public BigInteger nextBigInteger() throws IOException{return new BigInteger(next());}public void println() throws IOException {bw.newLine();}public void println(int[] arr) throws IOException{for (int value : arr) {bw.write(value + " ");}println();}public void println(int l, int r, int[] arr) throws IOException{for (int i = l; i <= r; i ++) {bw.write(arr[i] + " ");}println();}public void println(int a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(int a) throws IOException{bw.write(String.valueOf(a));}public void println(String a) throws IOException{bw.write(a);bw.newLine();}public void print(String a) throws IOException{bw.write(a);}public void println(long a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(long a) throws IOException{bw.write(String.valueOf(a));}public void println(double a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(double a) throws IOException{bw.write(String.valueOf(a));}public void print(char a) throws IOException{bw.write(String.valueOf(a));}public void println(char a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}}
}

E - Last Train

题目大意

$n$ 个点 $m$ 条边，每条边有信息 $l,d,k,c$
表示 $l,l+d,l+2*d,\cdots ,l+(k-1)*d$ 时刻有代价为 $c$ 的路径
问 $1\rightarrow n-1$ 每个点到 $n$ 的最长距离

解题思路

单一汇点，多源点，反向建图
若当前时间为 $tim$ ，则到下一个点的时间为 $may=tim-c$
则下一点最晚出发的时间为其等差数列中最大的小于 $may$ 的时刻
由于有最晚出发时间的限制，所以不会有走环的情况

import java.io.*;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.Vector;public class Main{static long md=(long)998244353;static long Linf=Long.MAX_VALUE/2;static int inf=Integer.MAX_VALUE/2;staticclass Edge{int fr,to,nxt;long l,d,k,c;public Edge(int u,int v,long L,long D,long K,long C) {fr=u;to=v;l=L;d=D;c=C;k=K;}}static Edge[] e;static int[] head;static int cnt;static void addEdge(int fr,int to,long l,long d,long k,long c) {cnt++;e[cnt]=new Edge(fr, to, l, d, k, c);e[cnt].nxt=head[fr];head[fr]=cnt;}static class Node{int x;long dis;public Node(int X,long D) {x=X;dis=D;}}public static void main(String[] args) throws IOException{AReader input=new AReader();PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));int n=input.nextInt();int m=input.nextInt();e=new Edge[m+1];head=new int[n+1];cnt=0;for(int i=1;i<=m;++i) {long l=input.nextLong();long d=input.nextLong();long k=input.nextLong();long c=input.nextLong();int u=input.nextInt();int v=input.nextInt();addEdge(v, u, l, d, k, c);}PriorityQueue<Node> q=new PriorityQueue<Node>((o1,o2)->{if(o2.dis-o1.dis>0)return 1;else if(o2.dis-o1.dis<0)return -1;else return 0;});long[] dis=new long[n+1];Arrays.fill(dis, -1);dis[n]=Linf;q.add(new Node(n, Linf));while(!q.isEmpty()) {Node now=q.peek();q.poll();int x=now.x;if(x!=n&&dis[x]>=now.dis)continue;dis[x]=now.dis;for(int i=head[x];i>0;i=e[i].nxt) {int v=e[i].to;long c=e[i].c;long may=dis[x]-c;long l=e[i].l;long d=e[i].d;long k=e[i].k;if(may>=l) {may=l+Math.min((may-l)/d, k-1)*d;q.add(new Node(v, may));}}}for(int i=1;i<n;++i) {if(dis[i]==-1) {out.println("Unreachable");}else out.println(dis[i]);}out.flush();out.close();}//System.out.println();//out.println();staticclass AReader{ BufferedReader bf;StringTokenizer st;BufferedWriter bw;public AReader(){bf=new BufferedReader(new InputStreamReader(System.in));st=new StringTokenizer("");bw=new BufferedWriter(new OutputStreamWriter(System.out));}public String nextLine() throws IOException{return bf.readLine();}public String next() throws IOException{while(!st.hasMoreTokens()){st=new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException{//确定下一个token只有一个字符的时候再用return next().charAt(0);}public int nextInt() throws IOException{return Integer.parseInt(next());}public long nextLong() throws IOException{return Long.parseLong(next());}public double nextDouble() throws IOException{return Double.parseDouble(next());}public float nextFloat() throws IOException{return Float.parseFloat(next());}public byte nextByte() throws IOException{return Byte.parseByte(next());}public short nextShort() throws IOException{return Short.parseShort(next());}public BigInteger nextBigInteger() throws IOException{return new BigInteger(next());}public void println() throws IOException {bw.newLine();}public void println(int[] arr) throws IOException{for (int value : arr) {bw.write(value + " ");}println();}public void println(int l, int r, int[] arr) throws IOException{for (int i = l; i <= r; i ++) {bw.write(arr[i] + " ");}println();}public void println(int a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(int a) throws IOException{bw.write(String.valueOf(a));}public void println(String a) throws IOException{bw.write(a);bw.newLine();}public void print(String a) throws IOException{bw.write(a);}public void println(long a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(long a) throws IOException{bw.write(String.valueOf(a));}public void println(double a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}public void print(double a) throws IOException{bw.write(String.valueOf(a));}public void print(char a) throws IOException{bw.write(String.valueOf(a));}public void println(char a) throws IOException{bw.write(String.valueOf(a));bw.newLine();}}
}

查看全文

http://www.shuangfujiaoyu.com/news/7806.html