2017 ip期末答案
2021/12/23 23:07:34
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Q1
a )
i ) declare a variable $ a $ of type integer and assigning an initial value of 6
ii )declare a variable initial_name of type character and assigning an initial value of 'Z'
iii)declare a pointer $ x $ of type double
iv)declare a function $ powr $ taking two integers as parameters and return a variable of type double
v)declare a three-dimensional arrays $ a $ of type double and the size is 10x6x8
b )
student number :the name of variable can't have the space
bob&alice: the name of variable can't have the sign of '&'
c)
i ) char x = 'a';
ii ) (++n)+3
iii ) int a[2] = {10};
iv ) for (i = 0 \(\color{red}{;}\) i < 5 \(\color{red}{;}\) i++);
v ) if (x \(\color{red}{==}\) 10) printf(‘%d’, x)
Q2
a )
scanf("%d %d %d",&a,&b,&c);
b )
printf("%.2f %.2f %.2f",a,b,c);
c )
i ) \(\color{red}{F}\) Because the grade is smaller than 60 so this equation $ grade >= 60 $ will return false,and it will go to $ printf("F\n") $
ii ) \(\color{red}{b = 1, c = 0}\) Beacuse the equation $ a==1 $ will return 0,that b is the same as before .But the sign behind the $ b=5 $ is \(\color{red}{;}\) .So the code $ c =0; $ will aslo run.Then c is 0.
d )
i ) \(\color{red}{4}\)
ii ) \(\color{red}{4.750000}\)
iii ) \(\color{red}{4.750000}\)
iv ) \(\color{red}{4.000000}\)
v ) \(\color{red}{3}\)
vi ) \(\color{red}{3.800000}\)
Q3
a)
i )
1. int main() 2.{ 3. int x; 4. for ( x = 1; x <= 10; x++ ) {// the x is from 1 go to 10 5. if ( x == 5 )//cheak whether x is 5 6. continue;//if x is 5 the go to the start of loop,which means don't output 5 7. printf( "%d ", x );.// out put x 8. } 9. return 0; 10. }
ii ) \(\color{red}{1{\quad}2{\quad} 3 {\quad}4{\quad}6{\quad}7{\quad}8{\quad}9{\quad}10}\)
b )
i )
1. int powr(int base, int n)// declare a function powr taking two integers as parameters and return a integer 3. int i, p=1; 4. for (i=1; i<=n; i++) p=p*base;// let p multiply base for n times,that we can get the base^n 5. return p; 6. } 7. int main() 8. { 9. int a = 4; 10. int b = 2; 11. printf("%d, %d\n", powr(3,b),powr(a,b));//use the fuction powr to calculate the 3^b and a^b 12. }
- the sign '^' in this means power but not nor
ii ) \(\color{red}{9,16}\)
c)
i )
if(!(x%3)) printf("it is divisible by number three");
ii )
int sum=0; for(int i=1;i<=10;++i){ if(!(i%3)) sum+=i; }
Q4
a )
int *p=(int *)malloc(20*640*480*8);
b )
\(\color{red}{0}\)
\(\color{red}{8765}\)
\(\color{red}{-987}\)
1. int reverse(int x) { 2. long long tmp = abs((long long)x);//get the absolute value of x 3. long long ret = 0; 4. while (tmp) {// get the tmp divided until it is 0 5. ret = ret * 10 + tmp % 10;//get the number of tmp,and plus the 10 6. if (ret > INT_MAX) return 0; //if the reserve number over integer maximum 7. tmp /= 10; 8. } 9. if (x > 0)//judge if it is postive 10. return (int)ret;//if it is postive 11. else 12. return (int)-ret;// if it is negative 13. } 14. int main (int){ 15. int x; 16. scanf ("%d\n", &x);//input x 17. printf ("%d\n",reverse (x)); 18. }
c )
1 void main() 2 { 3 int i = 0, j = 0, k = 0, x=9; 4 int arr[3][3][3]; 5 6 for(i=0;i<3;i++)//give a start number to arr from 9 7 { 8 for(j=0;j<3;j++) 9 { 10 for(k=0;k<3;k++) 11 { 12 arr[i][j][k] = x++; 13 } 14 } 15 } 16 17 for(i=0;i<3;i++)//reverse the all number of the array arr 18 { 18 for(j=0;j<3;j++) 19 { 20 for(k=0;k<3;k++) 21 { 22 arr[i][j][k] = reverse(arr[i][j][k]); 23 } 24 } 25 26 } 27 for(i=0;i<3;i++)//output the reverse number of array 28 { 29 for(j=0;j<3;j++) 30 { 31 for(k=0;k<3;k++) 32 { 33 printf("%d\t",arr[i][j][k]); 34 } 35 printf("\n"); 36 } 37 printf("\n"); 38 } 39 }
d )
9 1 11 21 31 41 51 61 71 81 91 2 12 22 32 42 52 62 72 82 92 3 13 23 33 43 53
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