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用）
- 对于0，特判（与%=0区分开）
java
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();
static
class 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
-
个点
条边，每条边有信息
-
表示
时刻有代价为
的路径 -
问
每个点到的最长距离
解题思路
- 单一汇点，多源点，反向建图
- 若当前时间为
，则到下一个点的时间为
- 则下一点最晚出发的时间为其等差数列中最大的小于
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
java
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;
static
class 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();
static
class 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();
}
}
}