質問編集履歴
3
test
CHANGED
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File without changes
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test
CHANGED
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@@ -20,7 +20,7 @@
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20
20
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21
21
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while(i<100){
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22
22
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23
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-
if ((i % 2) =
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23
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+
if ((i % 2) = 0)
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24
24
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25
25
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count++;
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@@ -32,7 +32,7 @@
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32
32
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33
33
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34
34
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35
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-
if((i % 2)=
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35
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+
if((i % 2)=0 && (i%3)=0){
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36
36
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37
37
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count++;
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38
38
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2
test
CHANGED
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File without changes
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test
CHANGED
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@@ -18,7 +18,7 @@
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18
18
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19
19
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20
20
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21
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-
while(i<100
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21
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+
while(i<100){
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22
22
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23
23
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if ((i % 2) == 1 )
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24
24
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@@ -28,7 +28,7 @@
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28
28
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29
29
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}
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30
30
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31
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-
printf("
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31
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+
printf("M 1=%d\n",count);
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32
32
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33
33
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34
34
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@@ -40,7 +40,7 @@
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40
40
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41
41
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}
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42
42
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43
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-
printf("
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43
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+
printf("M 2=%d\n",count);
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44
44
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45
45
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46
46
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1
誤字
test
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File without changes
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test
CHANGED
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@@ -3,3 +3,47 @@
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3
3
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１～１００までの数で、2の倍数の数の個数と２の倍数だが、3の倍数でない数の個数を
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4
4
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5
5
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出すにはどうすればよいですか。
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6
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+
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7
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+
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8
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+
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9
|
+
#include <stdio.h>
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|
10
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+
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11
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+
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12
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+
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|
13
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+
int main (void){
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14
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+
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15
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+
int i=1;
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16
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+
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|
17
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+
int count=0;
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18
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+
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19
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+
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|
20
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+
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21
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+
while(i<1000){
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22
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+
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|
23
|
+
if ((i % 2) == 1 )
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|
24
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+
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|
25
|
+
count++;
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26
|
+
|
|
27
|
+
i++;
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|
28
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+
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|
29
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+
}
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30
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+
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|
31
|
+
printf("Condition 1=%d\n",count);
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32
|
+
|
|
33
|
+
|
|
34
|
+
|
|
35
|
+
if((i % 2)==1 && (i%3)==0){
|
|
36
|
+
|
|
37
|
+
count++;
|
|
38
|
+
|
|
39
|
+
i++;
|
|
40
|
+
|
|
41
|
+
}
|
|
42
|
+
|
|
43
|
+
printf("Condition 2=%d\n",count);
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|
44
|
+
|
|
45
|
+
|
|
46
|
+
|
|
47
|
+
return 0;
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|
48
|
+
|
|
49
|
+
}
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