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Write a program that reads the number of students who passed and failed in the exams and displays the percentages. For example:

輸入及格與不及格人數
輸出及格與不及格百分數

Example input:

12 8

Example output:

Success Ratio: 60%
Fail Ratio: 40%
[Exercise] Coding (C) - asked in Chapter 4: Expressions by (12.1k points)
ID: 25730 - Available when: Unlimited - Due to: Unlimited

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86 Answers

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Hidden content!
#include <stdio.h>



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answered by (42 points)
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Hidden content!
#include <stdio.h>



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{
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answered by (42 points)
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Hidden content!
#include <stdio.h>
int main(void)
{
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answered by (-168 points)
0 0
-----------Re-judge-----------
Case 0: Correct output
Case 1: Correct output
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Hidden content!
#include <stdio.h>
int main ()
{
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answered by (-140 points)
0 0
-----------Re-judge-----------
Case 0: Correct output
Case 1: Correct output
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Hidden content!
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{
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answered by (-329 points)
0 0
-----------Re-judge-----------
Case 0: Correct output
Case 1: Correct output
0 like 0 dislike
Hidden content!
#include <stdio.h>

int main(void)

{
* * ** ***** * * a,b;
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answered by (-168 points)
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Hidden content!
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int a,b;

float c,d;
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}
answered by (-329 points)
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Hidden content!
#include <stdio.h>

int main(void)

{
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answered by (-168 points)
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Hidden content!
#include <stdio.h>

int main(void)

{
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}
answered by (-168 points)
0 like 0 dislike
Hidden content!
* * * * *** * ** **


int main(void)

{


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return 0;
}
answered by (-107 points)
0 0
prog.c: In function 'main':
prog.c:11:9: error: format '%d' expects argument of type 'int *', but argument 2 has type 'float *' [-Werror=format=]
 scanf("%d",&number_passes);
         ^
prog.c:12:10: error: format '%d' expects argument of type 'int *', but argument 2 has type 'float *' [-Werror=format=]
 scanf ("%d",&number_failures);
          ^
cc1: all warnings being treated as errors
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