Number of briquettes & Birthday Candles & building blocks

Number of coal balls

2016 The difficulty of the year title is obviously higher than 2017 It's much simpler in .

package lanqiao.lanqiao_2016;

/**
 * Created by ministrong on 2020/3/4.
 *
 *
  Number of coal balls 

  There's a bunch of coal balls , Pile up in a triangular pyramid . Specifically ：
  On the first floor 1 individual ,
  The second floor 3 individual （ In a triangle ）,
  The third level 6 individual （ In a triangle ）,
  The fourth level 10 individual （ In a triangle ）,
 ....
  If there is 100 layer , How many coal balls are there ？

  Please fill in the number of coal balls .
  Be careful ： You should submit an integer , Don't fill in any extra content or explanatory text .
 
  answer ：171700

 */
public class Q1_2016_7_meiqiu {

    public static void main(String[] args) {
        int res=0;// in total 
        int t=0;// Each layer 
        for(int i=1;i<101;i++){
            t=t+i;
            res+=t;
        }

        System.out.println(res);
    }
}

birthday candle

package lanqiao.lanqiao_2016;

/**
 * Created by ministrong on 2020/3/4.
 *
 *
  birthday candle 

  From a certain year on, you hold a birthday every year party, And blow out the same number of candles with the same age every time .

  Now count , He blew it out altogether 236 A candle .

  Excuse me, , How old was his birthday party Of ？

  Please fill in his birthday party The age of .
  Be careful ： You should submit an integer , Don't fill in any extra content or explanatory text .

 
 answer ：26
 */
public class Q2_2016_7_shengri {
    public static void main(String[] args) {

        for(int i=1;i<200;i++){
            int res=0;
            for(int j=i;j<200;j++){
                res+=j;
                if(res==236) System.out.println(i);
            }
        }
    }
}

  Building blocks

package lanqiao.lanqiao_2016;

/**
 * Created by ministrong on 2020/3/4.
 *
  Building blocks 

  Xiao Ming likes to build digital blocks recently ,
  Altogether 10 A building block , There's a number on each block ,0~9.

  Building block rules ：
  Each block is placed on top of the other two blocks , And it must be smaller than the two building block numbers below .
  Finally, it was built into 4 The pyramid of layers , All the building blocks have to be used up .

  Here are two qualified methods ：

 0
 1 2
 3 4 5
 6 7 8 9

 0
 3 1
 7 5 2
 9 8 6 4

  Please calculate the total number of such methods ？

  Please fill in the number indicating the total number .
  Be careful ： You should submit an integer , Don't fill in any extra content or explanatory text .

  answer ：768
 */

import java.lang.reflect.Array;
import java.util.Arrays;

/***
 *  Yes 1~9 Full Permutation , requirement ,a[0]<a[2] and a[3],a[1]<a[3] and a[4],a[2]<a[5] and a[6],a[3]<a[6] and a[7],a[4]<a[7] and a[8]
 */
public class Q3_2016_7_jimu {

    public static int[] a=new int[]{1,2,3,4,5,6,7,8,9};

    public static void main(String[] args) {

        int res=0;
        do{
            if(isok(a)) res++;

        }while(nextque());
        System.out.println(res);

    }

    public static boolean isok(int[] a){
        return a[0]<a[2] && a[0]<a[3] && a[1]<a[3] && a[1]<a[4] && a[2]<a[5] &&  a[2]<a[6] && a[3]<a[6] && a[3]<a[7] && a[4]<a[7] && a[4]<a[8];

    }
    public static boolean nextque(){
        int x=-1;
        for(int i=8;i>0;i--){
            if(a[i]>a[i-1]){
                x=i-1;
                break;
            }
        }
        if(x==-1) return false;
        int y=-1;
        for(int i=8;i>x;i--){
            if(a[i]>a[x]){
                y=i;
                break;
            }
        }
        int t=a[x];
        a[x]=a[y];
        a[y]=t;

        Arrays.sort(a,x+1,a.length);
        return true;
    }
}

Lexicographic order for complete arrangement ： With a[5]={35876} For example

1. Find the first decreasing adjacent number from right to left , Write down the subscript of the number on the left i： here 58 First decline , Record 5 The subscript of is i=1.

2. Find the first one from right to left a[i] Big numbers , Record its subscript j： here 6 It's the first one 5 Big , Record 6 The subscript j=4.

3. In exchange for a[i] and a[j] The number of , The sequence becomes ：36875.

4. Then subscript i The following sequence is arranged from small to large , Get the results ：36578.

ps： If the decreasing adjacent number is not found in the first step , It means that this arrangement is the last one .