Let n be a 4-bit integer whose 9 times is exactly its inverse ordinal number (the inverse ordinal number is the number formed by reversing the number of integers. For example, the inverse ordinal number of 1234 is 4321 Please write a program with VF, ask n, I want to write a program with Visual FoxPro!

Let n be a 4-bit integer whose 9 times is exactly its inverse ordinal number (the inverse ordinal number is the number formed by reversing the number of integers. For example, the inverse ordinal number of 1234 is 4321 Please write a program with VF, ask n, I want to write a program with Visual FoxPro!

（1089）
Let n = ABCD, according to the meaning of the question, abcdx9 = DBCA, first consider to determine the thousand digit a = 1 (otherwise 9 times of ABCD will not be four digits), then deduce d = 9; second consider the hundred digit multiplied by 9, after 9, there is no carry to the thousand digit, so we can know B = 0 or 1; after inspection, when B = 0, C = 8, meet the equation; when B = 1, the formula can not be established. So the four digit is 1089
51math

Design a program, input a four digit (integer), separate each digit, and output in reverse order. For example, input 1234, output 4321
I just learned C + +, there are many do not understand, please try to use the simplest sentences to tell me, otherwise I do not understand ah

#include
int main()
{
int n,m;
scanf("%d",&n);
m=0;
while (n>0)
{
m = m*10 + n%10;
n /= 10;
}
printf("%d\n",m);
return 0;
}
Input a positive integer, output its reverse order value

Design a C + + program, input a 4-digit, reverse order output. For example, 1234, output 4321

#include
using namespace std;
int reve_ int(int n){
int t=0;
while(n){
t*=10;
t+=n%10;
n/=10;
}
return t;
}
void main(){
int n;
cin>>n;
cout

Which four digit positive integers are k times (1

2178 * 4 = 8712

C language program: input ten numbers in turn and output the largest one

Wealth value!

C language input a number, judge the number is a few digits

Divide by ten until it's zero

Given the set a = {x | x square + X (P + 2) + 1 = 0} if the intersection r of a is equal to the empty set R {x belongs to R | x greater than zero}, the value range of P can be obtained

Because a intersection r equals an empty set R
That is, a is an empty set
That is, x ^ 2 + X (P + 2) + 1 = 0 has no real solution
So B ^ 2-4ac

Given that a = {x | x is less than or equal to 1}, B = {x | X's Square -- X is less than or equal to 0}, then a intersects B =?
A. {x | less than or equal to - 1} B, {x | - 1 less than or equal to x less than or equal to 0} C, {x | 0 less than or equal to x less than or equal to 1} D, {x | 0 less than or equal to 2}

Choose C
B=[0,1],A=(-∞,1],
So the intersection is C

The parabola y is equal to the square of minus 2 (x-4) - 2

A:
y=-2(x-4)^2 -2
The opening is downward and the axis of symmetry x = 4
Vertex (4, - 2)
Maximum - 2

It is known that X and y are opposite to each other, and the square of (x + 2) minus the square of (y + 2) is equal to 4, then how much is x and how much is y

x=-y
Substitute the square of (x + 2) minus the square of (y + 2) = 4
x+2)^2-(-x+2)^2=4
4*2x=4
x=0.5