APCS 2021/01/09 - A卷
===
###### tags: `2021 觀念題` `筆試題目`
```c=
int func ( int a[] , int s , int f) {
int min = 0, max = s - 1 ;
while ( min < max ){
int mid = (min + max) / 2;
if ( f > a[mid] ){
min = mid + 1;
} else if ( f < a[mid] ){
max = mid - 1;
} else {
return mid;
}
}
return -1;
}
int main(){
int a[8] = { 2, 3, 3, 5, 8, 8, 8, 8 } ;
printf("%d", func(a, 8, 8) ) ;
}
```
01. 上述輸出的答案為下列何者？
+ [A] 4
+ [B] 5
+ [C] 6
+ [D] 7
---
```c=
int main(){
char c = 'a';
switch(c){
case 'a':
printf("a");
case 'b':
printf("b");
case 'c':
printf("c");
default:
printf("d");
}
}
```
02. 上述輸出的答案為下列何者？
+ [A] a
+ [B] abc
+ [C] ad
+ [D] abcd
---
```c=
int level = 0;
int num[6] ;
for (int i = 1 ; i <= 6 ; i = i + 1 ) {
if ( level == 0 ) {
num[level] = i ;
level = level + 1 ;
} else if ( rand() % 2 == 0 ){
level = level - 1 ;
printf("%d ", num[level] ) ;
i = i - 1;
} else {
num[level] = i ;
level = level + 1 ;
}
}
while ( level ) {
level = level - 1 ;
printf("%d ", num[level] ) ;
}
```
03. rand() 是一個會回傳隨機整數的函數，試問下列選項何者是無法達成？
+ [A] 1 2 3 5 6 4
+ [B] 4 6 5 3 1 2
+ [C] 3 4 6 5 2 1
+ [D] 6 5 4 3 2 1
---
```c=
int f(){
int a[5] = { 9, 2, 4, 7, 3 } ;
int b[10] = {0, 1, 0, 1, 0, 1, 0, 1, 0, 1} ;
int c = 0;
for (int i = 0; i < 5 ; i = i + 1 ) {
c += b[ a[i] ];
}
return c;
}
```
04. 回傳的 c 值為多少？
+ [A] 1
+ [B] 2
+ [C] 3
+ [D] 4
---
```c=
int f ( int n ){
if ( n < 2 )
return n ;
else
return n + f(__________) ;
}
```
05. 執行上述的函數時回傳的結果： $f$(14) = 28 ，$f$(10) = 21 ， $f$(6) = 12，判斷式該如何選擇才能符合對應關係？
+ [A] $\frac{n+1}{2}$
+ [B] $\frac{n}{2}$
+ [C] $\frac{n-1}{2}$
+ [D] $\frac{n}{2}+1$
---
```c=
int f ( int n ){
int a[5] = { 5, 3, 10, 4, 20 } ;
int b[5] = {} ;
int size = 0 ;
for ( int i = 0 ; i < 5; i = i + 1 ) {
b[ size ] = a[i] ;
size = size + 1 ;
}
while ( size ) {
printf("%d ", b[ size - 1]);
size = size - 1 ;
}
}
```
06. 執行上述的函數時後輸出的結果？
+ [A] 3 4 5 10 20
+ [B] 5 3 10 4 20
+ [C] 20 10 5 4 3
+ [D] 20 4 10 3 5
---
```c=
for (int i = 0; i < 6 ; i += 1 ){
for (int j = 0 ; __________ ; j += 3 ){
printf("[%d]", i+j);
}
}
```
07. 執行為上述的程式碼後輸出結果為[1][2][3][4][7][5][8]，試問判斷是該如何選擇？
+ [A] $j < i$
+ [B] $j \leq i$
+ [C] $j > i$
+ [D] $j \geq i$
---
```c=
void reverse( int a[], int n ) {
for (int i = 0 ; __________ ; i += 1 ) {
int temp = a[i] ;
a[i] = a[ n-1-i ] ;
a[ n-1-i ] = temp ;
}
}
```
08. 為了讓陣列反向儲存時，判斷式應該如何選擇？
+ [A] $i < n$
+ [B] $i < \frac{n}{2}$
+ [C] $i \leq n$
+ [D] $i \leq \frac{n}{2}+1$
---
```c=
#define SIZE 7
int main() {
int outer = 0 , sum = 0 , index = 0 ;
int a[7] = { 2, 1, 3, 4, 5, 0, 6 } ;
while( outer < SIZE ){
for( index = 0 ; index < a[outer] ; index += 1 )
if( index > 5 )
break ;
else if ( a[ index ] >= a[ index + 1 ] ){
sum = sum + a[ index ] ;
a[ index ] = a[ index + 1 ] ;
}
outer = outer + sum ;
}
}
```
09. 執行完上述的程式碼後 sum 的值應該為多少？
+ [A] 0
+ [B] 2
+ [C] 3
+ [D] 6
---
```c=
void f(int x, int y){
int temp = x;
x = y;
y = temp;
}
int main(){
int x = 2, y = 5 ;
f(x, y) ;
printf("%d", (x - y) * (x + y + 4) );
}
```
10. 輸出數值為多少？
+ [A] 33
+ [B] 1
+ [C] -1
+ [D] -33
---
```c=
int f(int *a, int *b) ;
int g(int *a, int *b) ;
int main(){
int a = 32, b = 5 ;
int x = f(&a, &b) ;
printf("%d %d %d", x, a, b) ;
}
int f(int *a, int *b ) {
int c = 10 ;
*a = *a + *b / 5 ;
int p = g(b, a) % 2;
if (p % 2 == 0){
return *b - p ;
} else {
return *b + p ;
}
}
int g(int *a ,int *b ) {
return ( *a - *b ) * 5 + 2 ;
}
```
11. 輸出數值為多少？
+ [A] 5 33 5
+ [B] 5 32 5
+ [C]143 33 5
+ [D]143 32 5
---
```c=
int f(char s[], int n){
char t[1000] ;
int j = 0 ;
for (int i = 0 ; i < strlen(s) ; i += 1, j += 2) {
if (s[ i ] == '0'){
t[ j ] = 'w' ;
t[j+1] = 'a' ;
} else if ( _____ || _____ ) {
t[ j ] = 't' ;
t[j+1] = 'e' ;
} else if (s[ i ] == '3' || s[ i ] == '5'){
t[ j ] = 'd' ;
t[j+1] = 'o' ;
} else if (s[ i ] == '4' || s[ i ] == '6'){
t[ j ] = 'p' ;
t[j+1] = 'u' ;
} else {
t[ j ] = ' ';
j = j - 2 ;
}
}
t[j] = '\0';
printf("%s\n", t);
}
int main(){
char s[] = "0328675";
f(s, 8);
}
```
12. 輸出"wadoteputedo"時，判斷式應該如何選擇才符合題目要求？
原版是給定 3 個數字字串對應的英文字串...
+ [A] s[i] == '2' || s[i] == '7'
+ [B] s[i] == '2' || s[i] == '8'
+ [C] s[i] == '1' || s[i] == '8'
+ [D] s[i] == '1' || s[i] == '2'
---
```c=
int main(){
char s[] = "3*5/6+2-1" ;
char a[10] = {} ;
char b[10] = {} ;
int la = 0, lb = 0 ;
for ( int i = 0 ; i < strlen(s) ; i += 1 ){
switch( s[i] ){
case '+':
case '-':
case '*':
case '/':
a[la] = s[i] ;
la = la + 1 ;
break ;
default :
b[lb] = s[i] ;
lb = lb + 1 ;
}
}
while( la ){
la = la - 1 ;
b[ lb ] = a[ la ] ;
lb = lb + 1 ;
}
b[lb] = '\0' ;
printf("%s\n",b ) ;
}
```
13. 輸出的字串為何？
+ [A] 35621*/+ -
+ [B] */+ -35621
+ [C] 35621- +/*
+ [D] *35/9+2-1
---
```c=
int main(){
int n ;
scanf("%d", &n) ;
int mat[n][n] = {};
int x = n-1 , y = (n - 1) / 2 ;
for(int i = 1 ; i <= n * n ; i = i + 1 ){
int a = _____ ;
int b = _____ ;
if ( mat[a][b] ){
a = ( x + 1 )% n;
b = y ;
}
x = a ;
y = b ;
mat[x][y] = i ;
}
}
```
14. 邊長Ｎ為奇數的魔方陣，依照下列規則填入１到 $N^{2}$ 的數字就能保證每一行和每一列以及斜對角的數字總和都相同。
(1) 由最下面那一列的中間格開始填入1
(2) 接續格在前一格的左下方，除非超出邊界或格內已填數字
(3) 超出下邊界，則接續格改為左邊一行的最上面一列，除非超出左邊界如說明(6)
(4) 超出左邊界，則接續格改為下面一列的最右邊一行，除非超出下邊界如說明(6)
(5) 若接續格已填有數字，接續格改為原格上方一格
(6) 若下面及左邊同時超出邊界時，接續格改為原格的上方一格
+ [A] (x-1+n)%n , (y+1+n)%n
+ [B] (x-1+n)%n , (y-1+n)%n
+ [C] (x+1+n)%n , (y+1+n)%n
+ [D] (x+1+n)%n , (y-1+n)%n
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