CodeForces 264B Good Sequences dp

题目链接：点击打开链接

题意：

给定n个数，

构造一个序列（只能选给出的n个数，但数字可重复用）

使得序列严格递增且相邻的两个数字不互质

思路：

因为是严格递增，所以给输入的n个数排个序，相当于选n个数中的子序列了。

把每个数都分解质因数，然后用质因数转移方程即可。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Queue;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
public class Main {
	int[] prime = new int[N], len = new int[N]; int primenum;//有primenum个素数 math.h
	void PRIME(int Max_Prime){
		primenum=0;
		prime[primenum++]=2;
		for(int i=3;i<=Max_Prime;i+=2)
		for(int j=0;j<primenum;j++)
			if(i%prime[j]==0)break;
			else if(prime[j]>sqrt((double)i) || j==primenum-1)
			{
				prime[primenum++]=i;
				break;
			}
	}
	int n;
	int[] a = new int[N];
	ArrayList<Integer> G = new ArrayList();
	void go(int x){
		G.clear();
		for(int i = 0; prime[i]*prime[i]<= x; i++){
			if(x%prime[i] == 0){
				G.add(prime[i]);
				while(x%prime[i]==0)x/=prime[i];
			}
		}
		if(x>1)	G.add(x);
	}
	void work() throws Exception{
		n = Int();
		for(int i = 0; i < n; i++)a[i] = Int();
		for(int i = 0; i < N; i++)len[i] = 0;
		Arrays.sort(a, 0, n);
		PRIME(100000);
		for(int i = 0; i < n; i++){
			go(a[i]);
			int mx = 0;
			for(int j = 0; j < G.size(); j++){
				mx = max(len[G.get(j)] , mx);
			}
			for(int j = 0; j < G.size(); j++)
				len[G.get(j)] = mx+1;
		}
		int ans = 1;
		for(int i = 0; i < N; i++)ans = max(ans, len[i]);
		out.println(ans);
	}

    public static void main(String[] args) throws Exception{
        Main wo = new Main();
    	in = new BufferedReader(new InputStreamReader(System.in));
    	out = new PrintWriter(System.out);
  //  	in = new BufferedReader(new InputStreamReader(new FileInputStream(new File("input.txt"))));
  //  	out = new PrintWriter(new File("output.txt"));
        wo.work();
        out.close();
    }

	static int N = 100005;
	static int M = 101;
	DecimalFormat df=new DecimalFormat("0.0000");
	static int inf = (int)1e8;
	static long inf64 = (long) 1e18*2;
	static double eps = 1e-8;
	static double Pi = Math.PI;
	static int mod = (int)1e9 + 7 ;
	
	private String Next() throws Exception{
    	while (str == null || !str.hasMoreElements())
    	    str = new StringTokenizer(in.readLine());
    	return str.nextToken();
    }
    private int Int() throws Exception{
    	return Integer.parseInt(Next());
    }
    private long Long() throws Exception{
    	return Long.parseLong(Next());
    }
    private double Double() throws Exception{
    	return Double.parseDouble(Next());
    }
    StringTokenizer str;
    static Scanner cin = new Scanner(System.in);
    static BufferedReader in;
    static PrintWriter out;
  /*  
	class Edge{
		int from, to, dis, nex;
		Edge(){}
		Edge(int from, int to, int dis, int nex){
			this.from = from;
			this.to = to;
			this.dis = dis;
			this.nex = nex;
		}
	}
	Edge[] edge = new Edge[M<<1];
	int[] head = new int[N];
	int edgenum;
	void init_edge(){for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;}
	void add(int u, int v, int dis){
		edge[edgenum] = new Edge(u, v, dis, head[u]);
		head[u] = edgenum++;
	}/**/
	int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;
		int pos = r;
		r--;
		while (l <= r) {
			int mid = (l + r) >> 1;
			if (A[mid] <= val) {
				l = mid + 1;
			} else {
				pos = mid;
				r = mid - 1;
			}
		}
		return pos;
	}

	int Pow(int x, int y) {
		int ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}
	double Pow(double x, int y) {
		double ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}
	int Pow_Mod(int x, int y, int mod) {
		int ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			ans %= mod;
			y >>= 1;
			x = x * x;
			x %= mod;
		}
		return ans;
	}
	long Pow(long x, long y) {
		long ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}
	long Pow_Mod(long x, long y, long mod) {
		long ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			ans %= mod;
			y >>= 1;
			x = x * x;
			x %= mod;
		}
		return ans;
	}

	int gcd(int x, int y){
		if(x>y){int tmp = x; x = y; y = tmp;}
		while(x>0){
			y %= x;
			int tmp = x; x = y; y = tmp;
		}
		return y;
	}
	int max(int x, int y) {
		return x > y ? x : y;
	}

	int min(int x, int y) {
		return x < y ? x : y;
	}

	double max(double x, double y) {
		return x > y ? x : y;
	}

	double min(double x, double y) {
		return x < y ? x : y;
	}

	long max(long x, long y) {
		return x > y ? x : y;
	}

	long min(long x, long y) {
		return x < y ? x : y;
	}

	int abs(int x) {
		return x > 0 ? x : -x;
	}

	double abs(double x) {
		return x > 0 ? x : -x;
	}

	long abs(long x) {
		return x > 0 ? x : -x;
	}

	boolean zero(double x) {
		return abs(x) < eps;
	}
	double sin(double x){return Math.sin(x);}
	double cos(double x){return Math.cos(x);}
	double tan(double x){return Math.tan(x);}
	double sqrt(double x){return Math.sqrt(x);}
}

CodeForces 264B Good Sequences dp

原文：http://blog.csdn.net/qq574857122/article/details/44007659