标签：10 return int ip color 期末 quad 2017 red
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
标签：10,return,int,ip,color,期末,quad,2017,red 来源： https://www.cnblogs.com/donkey2603089141/p/15725577.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。