C 。。。。。。 Points, points, points

The program executes a nested loop, and at the beginning of the loop, the program executes for(i=0; i<2;i++) then execute for(j=0; j<3;j++) and then go down, at this time i=0, j=0, and the output result is 0 0

Then the program returns to for(j=0; j<3;j++), printf(%d %d) is executed again",i,j);i=0,j=1;Until the last time J=3, but at this time J is not less than 3, so the following printf(%d%d) is not executed",i,j);The loop then returns to for(i=0; i<2;i++), the value of i becomes 1, and then it becomes for(j=0; j<3;j++) loop, first initialize j to j=0 and go down until j=3 again, and finally i=2 when the whole loop is exited, and the program is executed!

main()

int i,j;

printf("i j");

for(i=0;i<2;i++)

for(j=0;j<3;j++)

printf(%d %d",i,j);

It's the same as yours in this way, it's more troublesome, but it's suitable for beginners and it's easier to understand.

A double cycle. Output the value of i,j and wrap the line each time.

1. Open the Python development tool Idle and create a new '.

2. F5 runs the program, list1 is sorted correctly, the purpose of writing this is to show that the binary search must be premised on an ordered list, if it is out of order at the beginning, it must be sorted first, when the amount of data is large, quick sort is a good choice, and then perform a dichotomy search.

3. With the idea of recursion, recursion must have an end condition.

4. if len(li)==1: li length is equal to 1, only compare this list element with the value to be found return li[0]==item.

5. If len(li)==0: The length of li is equal to 0, and there is still no value at the end of all searches. return false.

6. Add the main method to the program.

7. F5 runs the program, and prints out the dichotomy search result correctly, false true.

Take half until the final answer.

In general, for the function f(x), if there is a real number c, when x=c is f(c)=0, then x=c is called the zero point of the function f(x).

Solving the equation requires all zeros of f(x).

First find a and b, make f(a), f(b) different signs, indicating that there must be zero points in the interval (a, b), and then find f[(a+b) 2], now assume f(a)<0, f(b)>0, a if f[(a+b) 2]=0, the point is the zero point, if f[(a+b) 2]<0, then there is a zero point in the interval ((a+b) 2,b), according to the above method to find the function value of the midpoint of the interval, so that you can keep getting close to the zero point.

If f[(a+b) 2]>0, ditto.

By shrinking the interval where the zero point of f(x) is located by half each time, the two endpoints of the interval are gradually approaching the zero point of the function to obtain an approximate value of the zero point, which is called the dichotomy method.

Since the specific calculations of the calculation process are complex, but the way of each step is the same, it can be calculated by writing a program.

(1) and (2) are complete. However, the above binary lookup function is not well written.