1. Angry Birds (fill in the blanks)

Planet X Angry Birds like to hit the train!

A straight track, 1000 meters between the two trains.

The two trains (which may be called A and b) travel at 10 m/s per hour.

Angry Birds from a car, speed 50 m/s, crashed into B car,

Then go back to hit a car, and then back to hit B car, so reciprocating ....

The two trains are parked 1 metres apart.

Q: How many times has the angry bird hit the B car?

Note: You need to submit an integer (indicating the number of times you crashed the B car), do not fill in any other content.

1 Public classOne {2 //A, B's position A, B's size relative to A's starting point, the bird's speed is 50,a, B's speed all is ten3 Public Static Doublesum=0;//sum used for the number of collisions between birds and b during the specified period4 Public Static voidFunb (DoubleADoubleb) {//in the distance A, b birds hit B by a5 if(b-a<=1){6 System.out.println (sum);7 return;8 }9 DoubleTime= (b-a)/(50+10);//the time when the bird and B collided during the distance of A and BTenA=a+10*time;//the position of a after hitting OneB=b-10*time;//the position of B after the crash Asum++; -Funa (A, b);//The bird turns to a when the bird and B collide. - return; the } - - Public Static voidFuna (DoubleADoubleb) {//birds Fly by B to a - if(b-a<1) {//if the distance between the bird and the B is less than 1 after the collision, then sum-- +sum--; - System.out.println (sum); + } A DoubleTime= (b-a)/(50+10); atA=a+10*time;//the position of a after hitting -B=b-10*time;//the position of B after the crash - Funb (A, b); - return; - } - Public Static voidMain (String args[]) { in Doublea=0,b=1000; - Funb (A, b); to } +}

Angry Birds (two recursive functions)

The answer is: 9

2. Anti-Magic side (fill in the blanks)

The ancient books of our country record very early

2 9 4

7 5 3

6 1 8

This is a third-order magic side. Each row and the number on the diagonal are added equally.

Consider the opposite problem below.

It is not possible to fill nine Gongge with 1~9 numbers.

So that the numbers on each diagonal of each row are not equal to each other?

This should be able to do.

Like what:

9 1 2

8 4 3

7 5 6

Your task is to search all third-order anti-magic squares. and count the total number of species.

Rotate or mirror the same type.

Like what:

9 1 2

8 4 3

7 5 6

7 8 9

5 4 1

6 3 2

2 1 9

3 4 8

6 5 7

are counted as the same situation.

Please submit the number of third-order anti-magic side altogether. This is an integer, do not fill in any superfluous content.

1 Public classMain {2 Public Static intsum = 0;//counter3 4 Public Static voidFunintA[],intN) {5 if(n = = 9) {6 text (a);7 return;8}Else {9 for(inti = n; I < 9; i++) {//N Behind theTen inttemp =A[i]; OneA[i] =A[n]; AA[n] =temp; -Fun (A, n + 1); - intTamp =A[i]; theA[i] =A[n]; -A[n] =tamp; - } - } + } - + Public Static voidTextinta[]) { A intL1 = a[0] + a[1] + a[2]; at intL2 = A[3] + a[4] + a[5]; - intL3 = A[6] + a[7] + a[8]; - intS1 = a[0] + a[3] + a[6]; - intS2 = a[1] + a[4] + a[7]; - intS3 = A[2] + a[5] + a[8]; - intX1 = a[0] + a[4] + a[8]; in intx2 = a[2] + a[4] + a[6]; - if(L1! = L2 && L1! = L3 && L1! = S1 && L1! = S2 && L1! = S3 && L1! = x1 && L 1! =x2) to if(L2! = L3 && L2! = S1 && L2! = S2 && L2! = S3 && L2! = x1 && L2! =x2) + if(L3! = S1 && L3! = S2 && L3! = S3 && L3! = x1 && L3! =x2) - if(S1! = S2 && S1! = S3 && S1! = x1 && S1! =x2) the if(S2! = S3 && s2! = x1 && S2! =x2) * if(S3! = x1 && S3! =x2) $ if(X1! =x2)Panax Notoginsengsum++; - } the Public Static voidMain (string[] args) { + inta[]={1,2,3,4,5,6,7,8,9}; AFun (a,0); theSystem.out.print (SUM/8); + } -}

third-order anti-magic Square (Recursive)

The answer is: 3120

Seventh session (16) Blue Bridge Cup Java Group B final real title