For Loop, for Loop

For Loop, for Loop

Print the 9-9 multiplication table using nested for Loop

1*1 = 1

1*2 = 2 2*2 = 4

.........

1*9 = 1 2*9 = 18 ...... 9*9 = 81

 

Public class For_03 {

Public static void main (String [] args ){

For (int I = 1; I <= 9; ++ I ){
For (int j = 1; j <= I; ++ j ){
System. out. print (j + "*" + I + "=" + (I * j) + "");
}
System. out. println ();
}
}
}

(Operating principle) Comments: When I = 1, I <= 9, enter the for loop below; j = 1, j <= I, if the condition is met, the output is (j + "*" + I + "=" + (I * j) + ""), that is, the output is 1*1 = 1, and a space is added, after the output, + j is returned. At this time, + j = 2, judge whether 2 is <= I (1), 2> 1 does not meet the condition, so no j * I is output, output line feed. Enter ++ I, ++ I is equal to 2, 2 <= 9 and meet the conditions. Enter the for loop below, j = 1, j <= I (2 ), output 1x2 = 2 in compliance with the condition, enter ++ j, ++ j = <= 2 in compliance with the condition, output 2x2 = 4, then, + + j = 3> 2. If the condition is not met, a line feed is output, and so on. When I = 9, j = 9, j <= I ends.

Running effect:

 

 

 

1 ~ 100 sum of all odd and even numbers.

 

Public class jiou {

Public static void main (String [] args ){

Int j = 0; // declare a variable j with int and assign a value of 0. The variable declaration involves three steps: dividing the data type into memory space, naming, and assigning values.

For (int I = 1; I <= 100; I + = 2) {// here is the for loop, declare a variable I with the int data type, and assign a value of 1, so that the maximum I value cannot exceed 100, each cycle I adds itself to 2

// Three elements of a loop: Initial Value (I = 1, indicating that I is output from the beginning)

// Termination condition (that is, if I <100, exceeds 100, the cycle ends)

// Step size: (I + = 2. Every cycle I is added with 2, so the step size is 2)

J + = I;

}

System. out. println ("the odd number of 1-and" yes :");

System. out. println (j );

// Output the variable j. Set 1 + 3 = 4, 4 + 5 = 9 .... display, 9, 16 ..... that is, the value overwritten by each addition is always added to the last value of I = 99, that is, the odd number and 2500 in 1-

 

Int o = 0;
For (int I = 0; I <= 100; I + = 2 ){
O + = I;

}
System. out. println ("the even number of 1-and the sum is :");
System. out. println (o); // Same principle as above

Running effect: