Huawei pen Questions 3 answers

Write oneProgram, Required function: Obtain the sum of the three numbers, 1, 2, 5, and 100.

For example, the first 100 is a combination, and the fifth one and the 19th five are a combination .... Please use C ++.

 

Answer: The easiest thing to thinkAlgorithmYes:

Set X to the number of 1, Y to the number of 2, Z to the number of 5, and number to the combination.

Note that 0 <= x <=, 0 <= Y <= 50, 0 <= z = 20, so you can program it:

 

 

 
Number = 0; For (x = 0; X <= 100; X ++) for (y = 0; y <= 50; y ++) for (Z = 0; z <= 20; Z ++) if (x + 2 * Y + 5 * z) = 100) number ++; cout <number <Endl;
 

 

 

The above program has to cycle 100*50*20 times, very low

In fact, this question is an obvious mathematical problem, not simply a programming problem. The solution is as follows:

Because X + 2y + 5z = 100

So X + 2y = 100-5z, and z <= 20x <= 100 Y <= 50

So (x + 2y) <= 100, and (x + 5z) is an even number

Perform a loop on Z. The possible values of X are as follows:

 

Z = 0, x = 100, 98,96,... 0

Z = 1, x = 95, 93,..., 1

Z = 2, x = 90, 88,..., 0

Z = 3, x = 85, 83,..., 1

Z = 4, x = 80, 78,..., 0

......

Z = 19, x = 5, 3, 1

Z = 20, x = 0

 

Therefore, the total number of combinations is an even number less than 100 + an odd number less than 95 + an even number less than 90 + an odd number less than... + 5 + 1,

That is:

51 + 48 + 46 + 43 + ........ 1

 

 

The number of even numbers (including 0) within an even m can be expressed as M/2 + 1 = (m + 2)/2

An odd number within an odd m can also be expressed as (m + 2)/2

 

Therefore, the total number of combinations can be programmed as follows:

 
Number = 0; For (intm = 0; m <= 100; m + = 5) {number + = (m + 2)/2 ;} cout <number <Endl;
 

This program only needs to loop 21 times and two variables to get the answer, which is much more efficient than the above program.