先输入27然后选中Inv
按“x^y”键
输入3
然后按=键
可得3
/iknow-pic.cdn.bcebos.com/4d086e061d950a7b9972012806d162d9f3d3c993"target="_blank"title="点击查顷梁看大图"class="ikqb_img_alink">/iknow-pic.cdn.bcebos.com/4d086e061d950a7b9972012806d162d9f3d3c993?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto"esrc="https://iknow-pic.cdn.bcebos.com/4d086e061d950a7b9972012806d162d9f3d3c993"/>隐历
#include <stdio.h>int 迟敬main(){
int i,n,k=0,m=0,s=0
printf("please input a short type integer:")
scanf("%d",&n)
while(n)
{m=m*10+n%10s+=n%10k++n/=10}
for(i=0i<ki++){
printf("%d+",m%10)
m/=10
}
printf("\b=%d",s)
return 0
}
运行稿旦带示键芦例:
使用java的拿凯指话可以这孙差样实现:import java.util.Scanner
public class s{
public static void main(String[] args){
Scanner input=new Scanner(System.in)
System.out.println("请输入一个正消配整数:")
int n=input.nextInt()
String s=n+""
String tmp=""
int l=s.length()
int sum=0
for(int q=lq>=1q--){
int c=(int)(n/Math.pow(10,q-1))
n=(int)(n%Math.pow(10,q-1))
sum+=c
if(q!=1){
tmp+=c+"+"
}
else{
tmp+=c
}
}
System.out.println(tmp+"="+sum)
}
}
欢迎分享,转载请注明来源:内存溢出
评论列表(0条)