C语言 C程序打印给定数字的数字平方和？

C program to print sum of squares of digits of a given number?

提问by ThisSiteSucks

I want to write a c program that prints the sum of the squares of a given number. For example, if the given number is 456, the output would be 4^2+5^2+6^2=16+25+36=77.

我想编写一个 ac 程序来打印给定数字的平方和。例如，如果给定的数字是 456，则输出将是 4^2+5^2+6^2=16+25+36=77。

So i have written this code and i want to know why it doesn't work if the user gives a number like 100,101,102 or 200,300 etc. It works fine for other numbers. I guess it has something to do with the dowhile loop. Please help me.

所以我写了这段代码，我想知道为什么如果用户给出一个像 100,101,102 或 200,300 等的数字它不起作用。它适用于其他数字。我想这与 dowhile 循环有关。请帮我。

#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
 int n,t=0,r,q;
 printf("Enter the number to be tested: ");
 scanf("%d",&n);
 q=n;
 do
 {
      r=q%10;
      t=t+pow(r,2);
      q=q/10;
 }
 while(q%10!=0);
 printf("%d",t);
 getch();


}

回答by dasblinkenlight

Your stopping condition is wrong: q%10!=0will become "true" as soon as you reach the first zero in a decimal representation. For example, for a number 6540321your program would add 3²+2²+1², and stop, because the next digit happens to be zero. Squares of 6, 5, and 4 would never be added.

您的停止条件是错误的：q%10!=0只要您达到十进制表示中的第一个零，就会变为“真”。例如，对于一个数字，6540321您的程序将添加 3 ²+2 ²+1 ²，然后停止，因为下一位数字恰好为零。永远不会添加 6、5 和 4 的平方。

Use q != 0condition instead to fix this problem. In addition, consider replacing

改用q != 0条件来解决这个问题。另外，考虑更换

t=t+pow(r,2);

with a more concise and C-like

用更简洁和类似C的

t += r*r;

回答by Spikatrix

Change

改变

while(q%10!=0);

To

到

while(q);

Which is the short for

哪个是缩写

while(q!=0);

This is done to prevent to loop from ending once the value of qis a multiple of 10.

这样做是为了防止循环在值q是 10 的倍数时结束。