This is the 5th day of my participation in the August More Text Challenge

Title 1:

Use the while loop to print 1, 2, 3, 4, 5, 6, 8, 9, 10
count = 0 while count < 10: count+=1 if(count ! = 7) : print (count)Copy the code

Answer:

This is the easy one, just print the numbers from 1 to 10 except for 7

  1. Let’s make it a little bit simpler and print out a loop from 1 to 10

     count = 0 ​ while count <= 10: ​   print(count) ​   count += 1
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    Note: beginners should be careful not to drop count += 1 in order to make the program close

    2. Then remove the 7, do not print, as long as it is not 7 print, so add a conditional judgment

    if(count ! = 7): print(count)Copy the code

    3. The final code is

    count = 0 while count < 10: count+=1 if(count ! = 7) : print (count)Copy the code

Topic 2:

Take the sum of all numbers from 1 to 100
 sum = 0 ​ i = 1 ​ while i <= 100: ​   sum += i ​   i+=1 ​ print(sum)
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Answer:

1. They want to sum 1-100, so we can narrow it down a little bit. Let’s sum 1 and 2 first

 a = 1 b = 2 sum = a + b
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2. Let’s take the sum of 1-3 again

 a = 1 b = 2 c = 3 sum = a + b + c
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3. However, if we have a lot of numbers, we can’t assign them one by one, so we use a loop to do the assignment. We still use sum to record the sum

Sum = 0 # sum = 0 while I <= 3: # set the loop condition sum += I # sum = 1 # I = 1 # ICopy the code

Sum is like a big house, and I is like a porter, carrying numbers from 1 to 3 in and adding them up

4. Then we can change it to the sum of the numbers from 1 to 100 and print the result. The final code:

 i = 1 sum = 0 while i <= 100:     sum += i     i += 1 print(sum)
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Title 3:

Print all odd numbers between 1 and 100
i = 1 while i <= 100: if(i%2 ! • print(I) I +=1Copy the code

1. Let’s narrow it down and print numbers 1-5 first, in a loop

 i = 1 while i <= 5:     print(i)     i += 1
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The difference between an odd number and an even number is whether it can be integer by 2. In other words, if this number is divided by 2, it is even if the remainder is 0, and odd if the remainder is not 0. We only want odd, so we add a criterion

if(i%2 ! = 0): print(i)Copy the code

3. Then you can print odd numbers 1-5

i = 1 while i <= 5: if(i%2 ! = 0): print(i) i += 1Copy the code

4. Finally, just change the loop condition to within 1-100

i = 1 while i <= 100: if(i%2 ! = 0): print(i) i += 1Copy the code

Topic 4:

Print all even numbers from 1 to 100
I =1 while I <= 100: if(I %2 == 0): • print(I) I +=1Copy the code

So this is a very similar problem to the last one

Title 5:

For 1-2 + 3-4 + 5… The sum of all the numbers of 99

Plan a

 sum = 0 i = 1 while i <= 99:     if(i%2 == 0):         tmp = -i     else:         tmp = i        sum += tmp     i+=1 print(sum)
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Answer:

1. Let’s narrow down the range and calculate the numbers between 1 and 5. It’s not hard to see from observation that the even numbers here are preceded by a negative sign, so we should find all the even numbers through conditional judgment, and then add a negative sign to them, and then add them to sum

 if(i%2 == 0):     tmp = -i  else:      tmp = i
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Tip: The temporary TMP variable is used here so as not to affect the loop because I is the key variable in the loop. We only want positive or negative values of I, but if I is negative it affects the loop, so make sure I executes

2. Then you can add sum to sum

 if(i%2 == 0):     tmp = -i else:     tmp = i sum += tmp
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3. Finally put the loop on

 sum = 0 i = 1  while i <= 5:      if(i%2 == 0):           tmp = -i      else:           tmp = i      sum += tmp      
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4. Then change the loop condition to 1-99

 sum = 0 i = 1 while i <= 99:     if(i%2 == 0):         tmp = -i     else:         tmp = i        sum += tmp     i+=1 print(sum)
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Scheme 2

Answer:

We can sum them separately, take the odd sum first, and then take the even sum, because the even sum is negative, so the odd sum plus the even sum is the same thing as the odd sum minus the even sum

1. Let’s narrow down the range first, calculate 1-5 and define two variables, which are odd and even and respectively

 sum_odd = 0 sum_even = 0
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2. We need a condition to determine whether I is odd or even, and if it’s odd we add it to sum_odd, and if it’s even we add it to sum_even

if(i%2 ! = 0): sum_odd += i elif(i%2 == 0): sum_even += iCopy the code

3. Finally loop, then odd and – even sum, finally output the result

sum_odd = 0 sum_even = 0 i = 1 while i <= 5: if(i%2 ! = 0): sum_odd += i elif(i%2 == 0): sum_even += i i+=1 sum = sum_odd - sum_even print(sum)Copy the code

4. Finally change the loop condition, the final code:

sum_odd = 0 sum_even = 0 i = 1 while i <= 99: if(i%2 ! = 0): sum_odd += i elif(i%2 == 0): sum_even += i i+=1 sum = sum_odd - sum_even print(sum)Copy the code