if(sum == arms) printf("%d is an armstrong number.", arms); else printf("%d is not an armstrong number.", arms); return 0; } 输出(Output) 该方案的产出应该是 -
[C]Armstrong Number //1234 = 1^4 + 2^4 + 3^4 + 4^4;//341 = 3^3 + 4^3 + 1^3//类似于水仙花数#include<stdio.h>#include<math.h>intdigitCount(intn) {//此函数用于获取整数的位数inti =0;do{ n= n /10; i++; }while( n >0);returni; }intisArmstrong(intn) {intdigitNum...
Armstrong number in C with programming examples for beginners and professionals covering concepts, control statements, c array, c pointers, c structures, c union, c strings and more.
C实现Armstrong数 Armstrong数,也称为自恋数或水仙花数,是指一个n位数(n≥3),它的每个位上的数字的n次方之和等于它本身。例如,1^3 + 5^3 + 3^3 = 153。 在C语言中,我们可以编写程序来判断一个数是否是Armstrong数。以下是实现Armstrong数的C代码示例: #include <stdio.h> #include <math.h> int ...
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#include<iostream>usingnamespacestd;intpower(intc,intd);intmain() {intnum;intf,rem,sum=0,temp,a=0; cout<<"Enter any number :: "; cin>>num; temp=num;while(temp != 0) { temp=temp/10; a=a+1; } f=num;while(f!=0) { rem=f%10; sum = sum + power(rem,a); f=f/10; ...
checking whether the indexing variable is armstrong or not using a user defined function and printing it if it is. The user defined function works simply by using a while loop to isolate digits and matching the sum of the cubes to the original number. If it is true, then returns 1 otherw...
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