Input any 4-digit positive integer from the keyboard, program to separate the 4-digit digits, calculate their sum and output to the display #include int main() { int a,b,c,d,e,f; scanf("%d",&e); e=(1000*a)+(100*b)+(10*c)+(1*d); f=a+b+c+d; printf("%d\n",f); return 0; } What's wrong with that?

Input any 4-digit positive integer from the keyboard, program to separate the 4-digit digits, calculate their sum and output to the display #include int main() { int a,b,c,d,e,f; scanf("%d",&e); e=(1000*a)+(100*b)+(10*c)+(1*d); f=a+b+c+d; printf("%d\n",f); return 0; } What's wrong with that?

It seems that you don't understand the assignment sign of C language? E = (1000 * a) + (100 * b) + (10 * c) + (1 * d); in this way, the value of (1000 * a) + (100 * b) + (10 * c) + (1 * d) is assigned to e, instead of decomposing e into a, B, C, D. change it to this way: # includeint main () {int a, B, C, D, e, f; scanf (% d ", & E); a = E / 100

C language programming, primary ~ design algorithm, input a 4-bit positive integer, invert them, for example, input 1234, output 4321
Design algorithm, input a four digit positive integer, invert them, for example, input 1234, output 4321. (hint: apply / integer part and% remainder part respectively, for example, 7 / 5 = 1, 7% 5 = 2)
It's written in TC file. I hope it's more detailed, so I can type it directly. I can also write spaces and punctuation,

#include #include int main(){int oldnum;int newnum=0;int temp;printf("please input number\n");scanf("%d",&oldnum);printf("the old number is %d\n",oldnum);while (oldnum !=0){newnum = newnum*10+oldnum%1...

Use 1234 to form different four digit integers. What is the sum of these numbers? How to calculate?

Sum = 666601234 four digits 1234 1243 1324 1342 1423 143222134 2143 2314 2341 2413 2431... And so on, the sum of all the numbers, ten digits, hundreds and thousands, each is 1.2.3.4, these four numbers appear six times, 1x6 + 2x6 + 3x6 + 4x6 = 60, don't forget to carry, finally get 666

. C programming language: input a positive integer, count the number of zeros in each digit, and find the largest one in each digit

#include<stdio.h>int main(){int n,max=-1,s=0;scanf("%d",&n);while(n){if(n%10==0)s++;if(n%10>max)max=n%10;n/=10;}printf("0:%d\nmax=%d\n",s,max);return 0;}

On line first simplify and then evaluate 2 (A's square B + AB's Square) - 2 (A's Square b)
2 times (A's square B + AB's Square) - 2 (A's Square B-1) - 2Ab's square-2 a = - 2 b = 2
The square y of 3x - [the square of 2XY - 2 (XY - the square y of 3 / 2 x) + XY]

The parameter equation of the curve xy = 1 is?
A.x=t^(1/2) y=t^(1/2)
B.x=sinα y=1/sinα
C.x=cosα y=1/cosα
D.x=tanα y=1/tanα
What do you think of domain?

In B, sin α belongs to - 1 to 1
Cos α in C belongs to - 1 to 1
The range of T ^ (1 / 2) greater than or equal to 0 in a is not correct

1. If x ^ 2 + ax + B = (x + 2) (X-5), then a =?, B =? 2. (X-2) (x + 1) is the result of which polynomial factorization. 3. Factorization factor (3x + y) (2x-3y) - 2Y (2x-3y) 4. Calculate by factorization: (1) 9999 ^ 2 + 9999 (2) (5 ^ 99-5 ^ 98) / (5 ^ 97-5 ^ 96)

Answer: 1. (x + 2) (X-5) = x ^ 2-3x-10, so a = - 3, B = - 10.2. (X-2) (x + 1) = x ^ 2-x-2, which is the result of factorization of x ^ 2-x-2. 3. The original formula = (2x-3y) (3x + y-2y) = (2x-3y) (3x-y) 4. (1). The original formula = 9999 * (9999 + 1) = 999990000 (2). The original formula = [5 ^ 98 (5-1)] / [5 ^ 96 (5-1)] = 5 ^ 2 = 25

D is a point in the plane coordinate, and the coordinate of D point is obtained when the parallelogram is composed of ABCD four points
A（-1,0） B（3,0） C（0,-3）

D (2,3) or (4, - 3) or (- 4, - 3)

With the precise definition of limit, it is proved that the following limit limx & # 178; = 1 x approaches 1
I just learned advanced mathematics, but I don't know how to prove limx & # 178; = 1 x →∞

To prove by definition that limit is actually a form of writing is to draw a gourd according to the example
Certificate limit 0

It is known that the parabola passes through the origin O and a point a 40 on the x-axis. The vertex of the parabola is e, and its axis of symmetry and X-axis intersect at point D
The parabola is known to pass through the origin O and the point a on the x-axis a on the O and x-axis of the origin O and the origin o of the parabola is known; the parabola is known to pass through the origin O and the origin O and the x-axis of the origin O and the point a on the X-axis a (A &; 4 &; 4 &; 0 &; 4 &; 4 &; 4; 4 &; 0; 0 &; the vertex of the parabola is E &; its symmetric axis and its x-axis is intersection point D &; its symmetric axis and x-axis and its x-axis is intersection point d &; a straight line y = - y = - y = - 2x-2x-1; y = - y = - 2x-the first one is 57351; the second one is 57348; the third one is 57349; Find the value of M and the corresponding analytic formula of the parabola, which is a point on the parabola, if s △ ADP = s △ ADC &;

① Let the equation of parabola be y = ax & # 178; + BX + C
And the parabola passes through O (0,0) point a (4,0)
So C = 0, y = a (X-2) ² - 4A
The straight line y = 2x-1 passes through point B (- 2, m), so m = - 5 and point B is on the parabola
Parabola y = (- 5 / 12) * (X-2) & #178; + 5 / 3
② The coordinate of point C is (0, - 1)
Let the coordinates of p be (g, H) (to distinguish x, y)
According to the definition, s △ ADP = 1 / 2 * ad * I h i s △ ADC = 1 / 2 * ad * 1
To make the areas of two △ equal, I h i = 1
Separate discussion: when h = 1, g = 2 + [√ (8 / 5)] or 2 - [√ (8 / 5)]
When h = - 1, g = 2 + [√ (32 / 5)] or 2 - [√ (32 / 5)]
So there are four P points
（2+【√（8/5）】,1） （2-【√（8/5）】,1）
（2+【√（32/5）】,-1） (2-【√（32/5）】,-1)