c++ Sentinel循环，存储最小和最大数字

Question

这是我的程序:

#include <iostream>
using namespace std;

int main(){
    int num;
    int numtotal = 0;
    int numcount = 0;
    int big = 0;
    int low = 0;

    cout<<"enter number or 0 to exit"<<endl;
    cin>>num;
    while(num != 0){
        numtotal = numtotal + num;
        numcount++;
        big = num;
        low = num;

        cout<<"enter number or 0 to extit"<<endl;
        cin>>num;

        if(num < low){
            low = num;
        }
        else if(num > big){
            big = num;
        }
    }

    cout<<"total of numbers: "<<numtotal<<endl;
    cout<<"totoal of numbers entered: "<<numcount<<endl;
    cout<<"biggest number: "<<big<<endl;
    cout<<"lowest number: "<<low<<endl;
}

“低”和“大”输出始终是用户输入的最后两个数字，例如:

( 5, 4, 3, 2)

最大的数字:2

最小数字:0

我做错了什么？谢谢

Answer 1

如果用户在 while 循环中输入 0，则不需要将 big 和 low 设置为 num以达到您想要的结果。

否则，如果您不这样做，在用户仅输入正数的情况下，您的输出最小数字将始终为 0，我认为您不希望这样:

#include <iostream>
using namespace std;

int main() {
    int num;
    int total = 0;
    int count = 0;

    cout << "Enter number or 0 to exit" << endl;
    cin >> num;

    int biggest = num;
    int lowest = num;

    while (num != 0) {
        total += num;
        count++;
        cout << "Enter number or 0 to exit" << endl;
        cin >> num;
        if (num != 0) {
            if (num < lowest) {
                lowest = num;
            } else if (num > biggest) {
                biggest = num;
            }
        }
    }

    cout << "Total of numbers: " << total << endl;
    cout << "Amount of numbers entered: " << count << endl;
    cout << "Biggest number: " << biggest << endl;
    cout << "Lowest number: " << lowest << endl;

    return 0;
}

示例用法:

Enter number or 0 to exit
5
Enter number or 0 to exit
4
Enter number or 0 to exit
3
Enter number or 0 to exit
2
Enter number or 0 to exit
0
Total of numbers: 14
Amount of numbers entered: 4
Biggest number: 5
Lowest number: 2