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HUAWEI Programming Contest 2024（AtCoder Beginner Contest 342）

news 来源：原创 2024/5/8 15:17:44

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();}}
}