Triangle problem worst-case test case calculation

Input 3 It's an integer a、b and c As the three sides of a triangle , requirement a、b and c The following conditions must be met ：

1、 Integers
2、3 Number
3、 The side length is greater than or equal to 1, Less than or equal to 100
4、 The sum of any two sides is greater than the third
Output is 5 One of the three situations ：
If you don't do that 1、2、3, Then the program output is “ Input error ”.
If you don't do that 4, Then the program output is “ Non triangle ”.
If the three sides are equal , Then the program output is “ Equilateral triangle ”.
If exactly two sides are equal , Then the program output is “ an isosceles triangle ”.
If all three sides are not equal , Then the program output is “ General triangle ”.

Algorithm description

take a,b,c The three sides are regarded as vectors vec(a,b,c), The data to be tested is saved in the array a in , that The vector becomes vec(a[i],a[j],a[k]). In this case a Stored in is 1,2,50,99,100, therefore i、j、k The numbers that make up ijk It can be regarded as a pental number , Change this hexadecimal number from 000 Traversing 444 We can work out vec All results

#include<iostream>
using namespace std;

int add(int a[],int n){
    int j = 0;

    while(a[j]<5&& j<n){
        if(a[n-j-1]+1<5){
            a[n-j-1] +=1;
            return 1;
        }else{
            a[n-j-1] = 0;
            j++;
        }
    }

    return 0;

}

void print(int a[], int n){
    for(int i= 0;i<n;i++){
        printf("%d ",a[i]);
    }
    printf("\n");
}

int main(){
    int vec[6],count[6];
    int v;
    int j = 0,left,right;
    int a[5];
    cout<<" Please enter the number of variables : ";
    cin>>v;

    cout<<" Please enter the left boundary : ";
    cin>>left;

    cout<<" Please enter the right boundary : ";
    cin>>right;

    for(int i = 0;i<v;i++){
        count[i] = 0;
    }

    a[0] = left;
    a[1] = left+1;
    a[2] = (right+left)/2;
    a[3] = right-1;
    a[4] = right;

    do{
        for(int i=0;i<v;i++){
            vec[i] = a[count[i]];
        }
        print(vec,v);

    }while(add(count,v));


}

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