C 将整数转为英文单词输出,如123

以上函数中for(;n%10!=0;k++)n=n/10;这个位置就不对了,如果输入750,想想看会是什么结果!for(;m%10!=0;k--)也和上边的错误有异曲同工之妙,嘿嘿……以下是我刚写的一

#include<stdio.h>#include<stdlib.h>voidmain(){    char*Eng1[20]={"ze

#includevoidmain(){intx,a,b,c;scanf("%d",&x);a=x/100;b=x/10%10;c=x%10;x=c*100+b*10+a;printf("%d",x);

核心思想利用判断不断用小的数去替代大的数开始丨输入abc丨判断框a>b--是--判断框a>c--是----（否)丨（否)丨丨令a=b令a=c---------丨丨--输出a--结束判断框a>c--是-

#includevoidmain(void){unsignedinti;scanf("%d",&i);i&=0xf0;i/=16;printf("0x%02x",i);printf("\n");}//

不对,实质上这种情况该用char类型,并判断,因为int可输入多位,例如输入11,12,14你这种算法就不能反向输出

#include"stdio.h"#include"math.h"intmain(void){charinput[100];printf("Pleaseinputthearray(int):");ge

#include#includevoidmain(){char*p[10]={"zero","one","two","three","four","five","six","seven","eight

usingSystem;usingSystem.Collections.Generic;usingSystem.Linq;usingSystem.Text;namespacetest{classPro

main(){intx;scanf("%d",&x);x=(x%10)*100+(x/10%10)*10+(x/100)printf("%d\n",x);}

正面看（有字的面）7805：1脚：电源正输入（+12V）2脚：地（负极）3脚：输出正（+5V）.输入输出侧最好要加100uF的滤波!

没必要这么复杂,可参考如下：#includeintmain(){intn,total,i,a[10];printf("请输入一个整数：");scanf("%d",&n);total=0;while(n

选择法排序：#include"stdio.h"#defineN10voidmain(){inti,j,k,a[N],temp;printf("请输入10个数字：\n");for(i=0;i

#include#includeintmain(){inttemp;inta,b,c;scanf("%d%d%d",&a,&b,&c);if(a>b)//保证a中存放最大值{temp=b;b=a;a=

#include#includeintmain(){charwords[10][7]={"zero","one","two","three","four","five","six","serven",

#include <stdio.h>#include <stdlib.h>#define N 30int main(){&nb

#includevoidmain(){inta,b,c;scanf("%d%d%d",&a,&b,&c);printf("%d\n",a>b?((a>c)?(a):(c)):(b>c)?b:c);}再

你这个是什么网站啊?看到好多在网上提交答案#includevoidmain(){intop1,op2,result;charch;scanf("%d%c%d",&op1,&ch,&op2);resul

假设12.34在A1,在B1输入=IF(A1再问：����̸����⣬������ķ��������£�һ������ֿ��ԣ�����0�ľͲ����ˣ���û��ʲô�����ܽ����再答：=I

#include"stdio.h"voidmain(){intx,y,z,t;scanf("%d%d%d",&x,&y,&z);if(x