First class exercise

One 、 Single topic selection

2-1
Overloaded functions are selected based on when calling , The wrong is （D）.

A. The parameters of the function
B. Type of parameter
C. Function name
D. Type of function

2-2
The following requirements for defining overloaded functions ,（ C） It's wrong. .

A. The number of required parameters is different
B. At least one parameter of different type is required
C. The return value of the function is required to be different
D. The same number of parameters is required , Different parameter types

Two 、 Function questions

6-1 Maximum function overload

fraction 10
author Yang Jun
Company Sichuan Normal University
Write overloaded functions myMax Two integers can be found separately , Three integers , The maximum value of two doubles .

Function interface definition ：

Sample referee test procedure ：

/* Please fill in the answer here */

int main(){
cout<<myMax(3,4)<<endl;
cout<<myMax(3,4,5)<<endl;
cout<<myMax(4.3,3.4)<<endl;
}
sample input ：
nothing

sample output ：
Here is the corresponding output . for example ：

4
5
4.3

#include<iostream>
using namespace std;

int myMax(int a,int b){
    
    return a>b?a:b;
}
int myMax(int a, int b, int c){
    
    return a>b?(a>c?a:c):(b>c?b:c);
}
double myMax(double a,double b){
    
    return a>b?a:b;
}

3、 ... and 、 Programming questions

7-1 Calculate the area of the triangle

fraction 10
author Zhang Dehui
Company Xi'an University of Posts and Telecommunications
Enter three numbers from the keyboard , Used to represent the length of three sides of a triangle . If you can form a triangle, you will output the area of the triangle , Otherwise output No.

Input format :
Please write the length of the three sides of the triangle here , for example ：
3.1 4.2 5.3

Output format :
Please output the area of triangle here , for example ：

6.50661

sample input :
3.0 4.0 5.0
sample output :
6

#include <iostream>
#include <cmath>
using namespace std;

int main(){
    
    double a, b, c, p, s;
    cin >> a >> b >> c;
    if(a+b<=c || a+c<=b || b+c<=a){
    
        cout << "No" << endl;
        return 0;
    }
    p = (a+b+c)/2;
    s = sqrt(p*(p-a)*(p-b)*(p-c));
    cout << s << endl;
    return 0;
}