动态规划

 

import java.util.*;
import java.math.BigInteger;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        BigInteger[] dp = new BigInteger[n+1];
        if(n<=1){
            System.out.println(1);
            return;
        }
        if(n==1){
            System.out.println(2);
            return;
        }
        dp[1]=new BigInteger(String.valueOf(1));
        dp[2]=new BigInteger(String.valueOf(2));
        for(int i = 3;i<=n;i++){
            dp[i] = dp[i-2].add(dp[i-1]);
        }
        System.out.println(dp[n]);
    }
}

import java.util.*;
public class Main {
    public static void main(String[] args){//只能通过80%案例
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int dist = sc.nextInt();
        int[] pos = new int[n+1];
        for(int i =1;i<=n;i++){
            pos[i] = sc.nextInt();
        }
        long sum =1;
     
        for(int i = 4;i<=n;i++){
            int maxIndex = i;
            for(int j = i-1;j>0;j--){
                if(pos[i]-pos[j]<=dist){
                    maxIndex = j;
                }else{
                    break;
                }
            }

            if(i-maxIndex>=2){
                int len = i - maxIndex;               
                sum=(sum+(long)(len*(len-1))/2);
            }
        }
        System.out.println(sum%99997867); 
    }
}

 

class Solution {
    public int translateNum(int num) {       
        String str = String.valueOf(num);
        int n = str.length();
     
        int[] dp = new int[n+1];
        dp[0] = dp[1] = 1;
   
        for(int i = 2;i<=n;i++){
            int digit = Integer.parseInt(str.substring(i-2,i));
            if(digit<=25&&digit>=10){
                dp[i] = dp[i-2]+dp[i-1];
            }else{
                dp[i] = dp[i-1];
            }
        }
        return dp[n];

    }
}
 //方法2 递归求解
public int translateNum(int num) {       
        if(num<=9){
            return 1;
        }
        if(num%100<=25&&num%100>=10){
            return translateNum(num/100)+translateNum(num/10);
        }else{
            return translateNum(num/10);
        }
    }

 序列模式匹配

 方法一.暴力求解

import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            String[] arr = sc.nextLine().trim().split(" ");     
            int l=-1,r=-1,d = Integer.MAX_VALUE;
            for(int i = 0;i<=arr[0].length()-arr[1].length();i++){
                int index = 0;
                if(arr[1].charAt(index)==arr[0].charAt(i)){                    
                    int tl = i,tr=-1;
                    for(int j = i;j<arr[0].length();j++){
                        if(arr[1].charAt(index)==arr[0].charAt(j)){
                            index++;
                            if(index==arr[1].length()){
                                tr=j;
                                if(tr-tl<d){
                                    d = tr-tl;
                                    l = tl;
                                    r=tr;
                                }
                                break;
                            }                           
                           
                        }
                    }
                }
                
            }
            System.out.println(l+" "+r);
        }
    }
}

 

 

//回溯法 通过率只有70%
import java.util.*;
public class Main{
    static int  mark = 0;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for(int i = 0;i<arr.length;i++){
            arr[i]=sc.nextInt();
        }
        int price = sc.nextInt();
        helper(arr,new int[n],price);
        System.out.println(mark);
    }
    public static void helper(int[] arr,int[] flag,int price){
        if(mark==1||price<0){
            return;
        }
        if(price==0){
            mark = 1;
            return;
        }
        for(int i = 0;i<arr.length;i++){
            if(flag[i]==1){
                continue;
            } 
            flag[i]=1;
            price-=arr[i];        
            helper(arr,flag,price);
            price+=arr[i];
            flag[i]=0;
        }
    }
}
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] arr = new int[n];
        for(int i = 0;i<arr.length;i++){
            arr[i]=sc.nextInt();
        }
        int price = sc.nextInt();
        boolean[][] dp = new boolean[n][price+1];
        dp[0][0]=true;
        for(int j =1;j<=price;j++){
            if(arr[0]==j){
               dp[0][j]=true;
            }    
        }
        for(int i =1;i<n;i++){
            for(int j =0;j<=price;j++){
                dp[i][j]=dp[i-1][j];
                if(j>=arr[i]){
                    dp[i][j]=dp[i][j]||dp[i-1][j-arr[i]];
                }                 
            }
        }
        System.out.println(dp[n-1][price]?1:0);
    }
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值