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  Exercise 10   

  1. ..clause 1..: Basic knowledge
  2. ..clause 2..: Program readout
  3. Memo
  4. ..clause 4..: Description type
  5. ..clause 5..: Basic knowledge(example solution)
  6. ..clause 6..: Program readout(example solution)
  7. Memo(example solution)
  8. ..clause 8..: Description type(example solution)


[1] Basic knowledge


[  1-1  ]
What is called the repetition to which the condition is previously judged like the while sentence?

[  1-2  ]
What is called the repetition from which the condition is judged back like the do-while sentence?
Example solution

It returns to contents.


[2] Program readout


[  2-1  ]
For what display is the following program a program?
Judging from the content of processing and the variable identifier, answer.
 
#include <stdio.h> 

int main(void)  
{
	int year = 0;  
	double money = 10000;  
	
	while (money < 15000) { 
		year++; 
		money *= 1.01;  
	}
	
	printf("%d , %f\n",year , money);  

	return 0;
}
Example solution

It returns to contents.


Memo


[  3-1  ]
I want to make the program that inputs the point of the test.
However, because there is a point of the test only up to 0-100
Input it again when the rest is input.
Example solution

It returns to contents.


[4] Description type


[  4-1  ]
Explain why it to be suitable for the input check in the do-while sentence concisely.
Example solution

It returns to contents.


[5] Basic knowledge(example solution)


[  1-1  ]
Judgment ahead

[  1-2  ]
Post-judgment
Problem

It returns to contents.


[6] Program readout(example solution)


[  2-1  ]
When you entrust 10000 yen to the bank of 1% an interest rate year
Program that displays how many years to take for it to become 15000 yen.
Problem

It returns to contents.


Memo(example solution)


[  3-1  ]
 
#include <stdio.h> 

int main(void)  
{

	int score;  
	
	Point..input.
		scanf("%d",&score);  
	} while (score < 0 || score > 100);  
	
	Printf ("Input point several %d\n" score);  

	return 0;
}
Do as follows when you want to make the message displayed at the re-input.
It is distinguished whether there are 0 in variable score and it is the first input.
When 0 is input to score, the condition doesn't put it on because it comes off the loop.
 
#include <stdio.h> 

int main(void)  
{

	int score = 0;  
	
	"The point :   If (Score =   0) ..printf (.. input it within the range of 0-100. \n");  Printf ("Input the point");  
		scanf("%d",&score);  
	} while (score < 0 || score > 100);  
	
	Printf ("Input point several %d\n" score);  

	return 0;
}
Problem

It returns to contents.


[8] Description type(example solution)


[  4-1  ]
Because the do-while sentence is a post-judgment loop, it is executed once without fail in the beginning.
Because the situation of not inputting it doesn't happen.
Problem

It returns to contents.


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