Practical question 1
[problem description]
Program, realize the following functions:
(1) Define two one-dimensional arrays x, y, no more than 50 elements.
(2) Enter k integers from the keyboard into array x.
(3) Calculate the average value ave of data in x and the number of elements greater than the average value n and output.
(4) Copy the data in array x to array y, store the repeated data only once, and output the data in Y finally.
[input form]
Input two lines, the first line is the number of data input, the second line is several integers, each integer is followed by a space for separation;
[output form]
There are several integers in the first line, each integer is followed by a space, and the last space is followed by a new line; there are several integers in the second line, each integer is followed by a space, and the last space does not need to be followed by a new line;
[sample input]
6 6 3 4 3 2 9
[sample output]
4.5 2 6 3 4 2 9
[Abstract]
#include<iostream> using namespace std; int main() { int x[50],y[50]; int i,j,sum=0,n=0,m=0; float avg=0; int k; cin>>k; for(i=0;i<k;i++) { cin>>x[i]; sum+=x[i]; } avg=(float)sum/k; for(j=0;j<k;j++) { if(x[j]>avg) n++; } for(i=1,y[0]=x[0],m=1;i!=k;i++) { for(j=0;j<i;j++) { if(x[i]==x[j]) break; } if(j==i) { y[m]=x[i]; m++; } } cout<<avg<<" "<<n<<endl; for(i=0;i<m;i++) cout<<y[i]<<" "; }
Practical question 2
[problem description]
There are 12 people sitting in a circle to play the counting game. The number of people who report to k will be eliminated from the game starting from the number 1 person. Then the number of people who report to k will be eliminated from the next person who report to k from the number 1 person who still reports to k from the next person who is eliminated from the game in a clockwise direction. This continues until only one person is left. Please write a program to output the number of the last person left.
Be careful:
(1) Suppose that the number of the participants is 1 to 12 in a clockwise direction, and the array can be used to store the data;
(2) K > 1, specified by the user through cin input.
[input form]
Enter an integer to represent the reported value;
[output form]
Output an integer, that is, the number of the last remaining person;
[sample input]
3
[sample output]
10
[Abstract]
#include <iostream> using namespace std; int main() { int k; int i=0,n=12,t=0; cin>>k; int a[13]; for(i=0;i<13;i++) { a[i]=i; } for(i=0,t=1;i<=k;t++) { if(t>=13) t=1; if(a[t]==0) continue; i++; if(i==k) { a[t]=0; n--; i=0; } if(n==1) break; } for(i=1;i<13;i++) if(a[i]) cout<<a[i]<<endl; }
Practical question 3
[problem description]
Xiao Zong wants to know the number of days between two dates. He wants to have a date calculator, which can automatically calculate the number of days between two dates after inputting them.
Requirements: design the corresponding function to calculate the completion days, and verify the correctness in the main function.
[input form]
Enter two dates in the order of MM DD YY, separated by a space;
[output form]
Output the number of days between two dates, i.e. an integer, after which no line wrapping is required;
[sample input]
2016 3 6 2017 1 1
[sample output]
301
[Abstract]
#include<iostream> #include<cmath> using namespace std; bool fun(int year) { if((year%4==0)&&(year%100!=0)) return true; else return false; } long int past(int year) { int p=0,q=0,i; for(i=0;i<year;i++) if(fun(i)) p++; else q++; return (366*p+365*q); } int now(int year,int month,int day) { int ans=0; if(fun(year)) { switch(month) { case 1:ans=day;break; case 2:ans=31+day;break; case 3:ans=31+29+day;break; case 4:ans=31+29+31+day;break; case 5:ans=31+29+31+30+day;break; case 6:ans=31+29+31+30+31+day;break; case 7:ans=31+29+31+30+31+30+day;break; case 8:ans=31+29+31+30+31+30+31+day;break; case 9:ans=31+29+31+30+31+30+31+31+day;break; case 10:ans=31+29+31+30+31+30+31+31+30+day;break; case 11:ans=31+29+31+30+31+30+31+31+30+31+day;break; case 12:ans=31+29+31+30+31+30+31+31+30+31+30+day;break; } } else { switch(month) { case 1:ans=day;break; case 2:ans=31+day;break; case 3:ans=31+28+day;break; case 4:ans=31+28+31+day;break; case 5:ans=31+28+31+30+day;break; case 6:ans=31+28+31+30+31+day;break; case 7:ans=31+28+31+30+31+30+day;break; case 8:ans=31+28+31+30+31+30+31+day;break; case 9:ans=31+28+31+30+31+30+31+31+day;break; case 10:ans=31+28+31+30+31+30+31+31+30+day;break; case 11:ans=31+28+31+30+31+30+31+31+30+31+day;break; case 12:ans=31+28+31+30+31+30+31+31+30+31+30+day;break; } } return ans; } int main() { int a,b,c,p,q,r; cin>>a>>b>>c>>p>>q>>r; cout<<abs(past(a)+now(a,b,c)-past(p)-now(p,q,r)); }
Practical question 4
[problem description]
For integer arrays a[10] and b[10], the following tasks are accomplished by programming:
(1) Input values for two arrays from the keyboard by the user;
(2) Find the maximum and minimum values of two arrays;
(3) The integers in arrays a and b are sorted from small to large and from large to small respectively;
(4) Make two ordered arrays a and b into an ordered array c[20] with a length of 20, so that the order of array c is from small to large
[input form]
Input two lines of integers, 10 for each line. The first line is the array in array a, and the second line is the value in array b;
[output form]
Output five lines, the first line has two integers, which are the maximum and minimum values of array a, separated by a space; the second line has two integers, which are the maximum and minimum values of array b, separated by a space; the third line outputs the values of array a in the order of small to large, with a space after each number, and the last one There are also spaces after the numbers; the fourth line outputs the values in array b in the order of large to small, with a space after each number and a space after the last number; the fifth line outputs the values in the combined array c in the order of small to large, with a space after each number and a space after the last number.
[sample input]
2 5 9 1 3 4 0 6 7 8 10 5 25 9 6 3 7 1 2 13
[sample output]
9 0 25 1 0 1 2 3 4 5 6 7 8 9 25 13 10 9 7 6 5 3 2 1 0 1 1 2 2 3 3 4 5 5 6 6 7 7 8 9 9 10 13 25
[Abstract]
#include<iostream> using namespace std; int comparemax(int x[]) { int max=0,i; for(i=0;i<10;i++) { if(max<x[i]) max=x[i]; } return max; } int comparemin(int x[]) { int min,i; for(i=0,min=x[0];i<10;i++) { if(min>x[i]) min=x[i]; } return min; } void up(int a[],int n) { int i,j,t=0; for(i=0;i<n-1;i++) { for(j=0;j<n-1-i;j++) { if(a[j]>a[j+1]) { t=a[j]; a[j]=a[j+1]; a[j+1]=t; } } } } void down(int a[]) { int i,j,t=0; for(i=0;i<10-1;i++) { for(j=0;j<10-1-i;j++) { if(a[j]<a[j+1]) { t=a[j]; a[j]=a[j+1]; a[j+1]=t; } } } } int main() { int a[10],b[10],i,c[20]; for(i=0;i<10;i++) { cin>>a[i]; } for(i=0;i<10;i++) { cin>>b[i]; } up(a,10); down(b); cout<<comparemax(a)<<" "; cout<<comparemin(a)<<endl; cout<<comparemax(b)<<" "; cout<<comparemin(b)<<endl; for(i=0;i<10;i++) { cout<<a[i]<<" "; } cout<<endl; for(i=0;i<10;i++) cout<<b[i]<<" "; cout<<endl; for(i=0;i<10;i++) c[i]=a[i]; for(i=10;i<20;i++) c[i]=b[i-10]; up(c,20); for(i=0;i<20;i++) { cout<<c[i]<<" "; } }
Practical question 5
[problem description]
Use the function of cin.getline() to input an English sentence from the keyboard, in which each word is separated by a space, and finally ends with '.'. Count the number of words in the sentence, and output each word in turn. The output number is followed by a new line, and each word is followed by a new line. Note: in this platform, the cin.getline() function is used as follows:
char s[50];
cin.getline(s,50); / / store up to 49 characters
[input form]
Enter an English sentence, in which each word is separated by a space, and the last word is followed by an English "." as the end;
[output form]
Output each word followed by a new line, and the last line outputs the number of words.
[sample input]
I like juice.
[sample output]
I like juice 3
(please input and output in strict accordance with the specified format.)
[Abstract]
#include<iostream> using namespace std; int main() { int i,num=0; char s[50]; cin.getline(s,50); for(i=0;;i++) { if(s[i]!=32&&s[i]!=46) cout<<s[i]; else { num++; cout<<endl; } if(s[i]=='.') break; } cout<<num; }
