用于循环分隔数字 [英] For loop divide numbers

问题描述

当谈到C ++时，我是一个业余爱好者，但是我已经接受了一个超过我的知识的任务。

任务是输入数字n，m。程序必须把它作为一个时间间隔，在这个时间间隔里，它检查是否有任何数字是指数相同的数字的总和。

编辑：（18.10 .15）
原来我没有正确理解我的任务。这里是：

用户输入两个数字，程序把它作为它检查所有数字的间隔，如果有一个数字在间隔中，

例如，我输入100和200.在这个区间里有153.
153 = 1 ^ 3 + 5 ^ 3 + 3 ^ 3（1 + 125 + 27）
节目显示为153元。 ;> N;
cin>>米;
for（int i = n; i <= m; i ++）
{
for（int k = n; k< = i; k ++）
{
a = n％10; //例如，我输入153，然后a = 3
f = n / = 10; // f = 15
b = f％10; // b = 5
f = f / = 10; // f = 1
c = f％10; // c = 1
f = f / = 10;
d = f％10;

（int j = 1; j <= 5; j ++）
{
a = a * a;
b = b * b;
c = c * c;
d = d * d;
if（a + b + c + d == n）
{
cout< n<< ENDL;

帮助将不胜感激。

解决方案

任务是输入数字n，m。程序必须把它作为一个时间间隔，在这个时间间隔里，它检查是否有任何数字是指数相同的数字的总和。

假设范围是 [n，m），那么这里是你的程序：

  return（n！= m）; 
  

任何数字都可以看作是指数相同的数字之和。例如：

  0 = 0 ^ 3 + 0 ^ 3 + 0 ^ 3 
 1 = 1 ^ 3 + 0 ^ 3 
 2 = 1 ^ 3 + 1 ^ 3 
 3 = 1 ^ 3 + 1 ^ 3 + 1 ^ 3 
  

等等。即使是负数也是如此。

所以在任何非空范围内至少有一个这样的数字。

I'm an amateur when it comes to C++ but I've already received a task which is over my knowledge.

Task is to enter numbers n,m. Programme must take it as an interval, in which it checks if there is any number which is a sum of numbers with the same exponent.

EDIT:(18.10.15) Turns out I didn't understood my task correctly. Here it is:

"User enter two numbers. Programme takes it as the interval in which it checks all the numbers. If there's a number in interval which all digit's sum of SAME exponent is that number, then programme shows it."

For example, I enter 100 and 200. In this interval there's 153. 153 = 1^3 + 5^3 + 3^3 (1+125+27) Programme shows 153.

cin >> n;
cin >> m;
    for (int i=n; i<=m; i++)
    {
        for (int k=n; k<=i; k++)
        {
                a = n % 10; //for example, I enter 153, then a=3
                f = n /= 10; //f=15
                b = f % 10;  //b=5
                f = f /= 10; //f=1
                c = f % 10; //c=1
                f = f /= 10;
                d = f % 10;

                for (int j=1; j<=5; j++)
                {
                    a = a * a;
                    b = b * b;
                    c = c * c;
                    d = d * d;
                    if (a + b + c + d == n)
                    {
                        cout << n << endl;
                    }
                }
        }
    }

Any help will be appreciated.

解决方案

Task is to enter numbers n,m. Programme must take it as an interval, in which it checks if there is any number which is a sum of numbers with the same exponent.

Assuming the range is given as [n, m), then here's your program:

return (n != m);

Any number can be seen as a sum of numbers with the same exponent. For example:

0 = 0 ^ 3 + 0 ^ 3 + 0 ^ 3
1 = 1 ^ 3 + 0 ^ 3
2 = 1 ^ 3 + 1 ^ 3
3 = 1 ^ 3 + 1 ^ 3 + 1 ^ 3

and so on. This is true even for negative numbers.

So in any non-empty range there exists at least 1 such number.

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