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问题是：编写一个函数，作为输入参数接收一个三位数的正数，因此必须得到相同的 3 位数除以中位数获得的最大和最小数之和。示例：函数 438 的输入参数 相同位数最大的是 843，最小的是 348，所以应该计算 (843 + 348) / 4。

我已经尝试过了，结果还可以，但是我的代码似乎很复杂，所以我问有没有更好的方法来做到这一点？

提前致谢

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

int check(int x) {

    int a, b, c, biggestNum, smallestNum, medianNum;

    a = x / 100;
    b = (x / 10) % 10;
    c = x % 10;

    if (a > b && a > c && b > c) {

        biggestNum= a * 100 + b * 10 + c;
        smallestNum= c * 100 + b * 10 + a;
        medianNum= b;

    }
    else if (a > b && a > c && b < c) {

        biggestNum= a * 100 + c * 10 + b;
        smallestNum= b * 100 + c * 10 + a;
        medianNum= c;

    }
    else if (b > a && b > c && a < c) {

        biggestNum= b * 100 + c * 10 + a;
        smallestNum= a * 100 + c * 10 + b;
        medianNum= c;

    }
    else if (b > a && b > c && a > c) {

        biggestNum= b * 100 + a * 10 + c;
        smallestNum= c * 100 + a * 10 + b;
        medianNum= a;

    }
    else if (c > a && c > b && a > b) {

        biggestNum= c * 100 + a * 10 + b;
        smallestNum= b * 100 + a * 10 + c;
        medianNum= a;

    }
    else if (c > a && c > b && a < b) {

        biggestNum= c * 100 + b * 10 + a;
        smallestNum= a * 100 + b * 10 + c;
        medianNum= b;

    }   

    cout << "Smallest number is: " << smallestNum<< " ,biggest is: " << biggestNum << " and median is: " << medianNum<< "." << endl;

    return (biggestNum + smallestNum) / medianNum;
}


int main() {

    cout << "Enter one 3 digit positive number: ";
    int x;
    cin >> x;

    float result = check(x);
    cout << "The result is: " << result << "." << endl;

    system("pause");
    return 0;
}
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4 回答 4

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考虑到结果是用整数算术计算的，发布的代码不能真正产生正确的答案：

int check(int x)  // <- note the type of the returned value
{
    int biggestNum, smallestNum, medianNum;
    // ...
    return (biggestNum + smallestNum) / medianNum;  // <- This is an integer division
}

int main()
{
    int x; 
    // ...   
    float result = check(x);  // Now it's too late to get the right result
}

该逻辑也没有考虑所有可能的情况，事实上它忽略了重复的数字，并且if else if缺少默认分支（最终无条件else）的大构造使那些未初始化的变量未确定，因此以下操作给出了毫无意义的结果.

鉴于分配限制，我会写如下内容

#include <utility>

// The assignment is about 3-digit numbers, you should check that x is actually in
// the range [100, 999]. Note that one of the extremes is a special case.
// Well, both, actually.
double I_ve_no_idea_how_to_name_this(int x)
{
    constexpr int base = 10;
    int smallest = x % base;
    x /= base;
    int median = x % base;
    x /= base; 
    // Note that this "works" (extracting the third digit) even if
    // x isn't a 3-digit number. If you can assure the input is well
    // defined, you can simplify this.
    int biggest = x % base; 

    // Now we can sort the previous variables.
    using std::swap;
    if ( median < smallest ) {
        swap(median, smallest);
    }
    // Now I know that smallest <= median
    if ( biggest < median ) {
        swap(biggest, median);
    }
    // Now I know that median <= biggest
    // ...
    // Is that enough or am I missing something here?
    // Please think about it before running the code and test it.

    // Once the variables are sorted, the result is easily calculated
    return (biggest + smallest + base * (2 * median + base * (biggest + smallest)))
           / static_cast<double>(median);
}
于 2019-11-17T19:42:29.167 回答
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尝试这个...

int check(int x) {
    int a,b,c,temp;
    a = x/100;
    b = (x/10)%10;
    c = x%10;
    if(b>a){
        temp=a;
        a=b;
        b=temp;
    }
    if(c>b){
        temp=b;
        b=c;
        c=temp;
    }
    if(b>a){
        temp=a;
        a=b;
        b=temp;
    }
    cout << "smallest: " << a+(b*10)+(c*100) << "\n";
    cout << "biggest: " << (a*100)+(b*10)+c << "\n";
    cout << "median: " << b << "\n";
    return (((a+c)*100)+(2*b*10)+(a+c))/b;
}
于 2019-11-17T23:27:07.450 回答
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首先，您应该使用更具描述性的变量名称，并且应该在定义时初始化每个变量。这两个步骤极大地有助于消除复杂程序中的错误。我知道这并不复杂，但这是一个好习惯。其次，标准库可以帮助找到最大和最小的数字，从而使其余的变得简单。所以这是一个没有任何if陈述的例子。

最后，使用命名空间标准；是一种不好的做法，应该避免。

double check(int x)
{
    int a = x / 100;
    int b = (x / 10) % 10;
    int c = x % 10;
    int bigdigit = std::max({ a, b, c }); // find largest
    int smalldigit = std::min({ a, b, c }); //find smallest
    int middledigit = a + b + c - bigdigit - smalldigit; // sum of all digits minus largest and smallest gives the remaining one
    int biggest = smalldigit + middledigit * 10 + bigdigit * 100;
    int smallest = smalldigit * 100 + middledigit * 10 + bigdigit;
    std::cout << "biggest: " << biggest << '\n';
    std::cout << "smallest: " << smallest << '\n';
    std::cout << "median: " << middledigit << '\n';
    return (1.0 * biggest + 1.0 * smallest) / (1.0 * middledigit); --using double instead of int, as result could be fractional
}
于 2019-11-17T23:46:32.597 回答
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检查此检查功能。

int check(int x) {

    if(x >= 1000) x %= 1000;  //or return -1;

    //get digits
    int M = x/100;
    int C = (x/10)%10;
    int m = x%10;

    //unrolled bubble sort.
    if(M < C) swap(M,C);
    if(C < m) swap(C,m);
    if(M < C) swap(M,C);

    //simplified formula
    return ((m+M)*(101))/C + 20;

}

//derivation of formula
B = M*100 + C*10 + m;
s = m*100 + C*10 + M;
B+s = (m+M)*100 + C*20 + (m+M)
    = (m+M)*(100 + 1) + C*20
(B+s)/C = ((m+M)*(100 + 1) + C*20)/C
        = ((m+M)*(101))/C + 20
于 2019-11-18T00:55:50.353 回答