C语言输入一个任意位数的正整数(不超过int范围),求其各位数的和
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#include"stdio.h"intmain(){\x09inti,j,n;\x09inta[12];\x09intmin,mx;\x09scanf("%d",&n);\x09for(i=0;i
main(){longinta;intb,c,d,e,f;scanf("%ld",&a);while(a>=1&&a=1&&a/10000=1&&a/1000=1&&a/100=1&&a/10=1&&
#includeintmain(){intnum=0;ints=0;inti=0;scanf("%d",&num);while(num){s+=num%10;num=num/10;i++;}print
#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}
#include"stdio.h"voidprt_diamond(intn){inti,j;for(i=1;i
#includemain(){intn,i,s=0,sum=0;printf("请输入一个整数:");scanf("%d",&n);i=n;while(i!=0){s+=(i%10)*(i%10)*(
#includeintmain(){intn;charch;inti;printf("输入一个10以内的正整数和相应个数的五进制成绩:\n");scanf("%d",&n);for(i=0
你写的实在是太繁琐了.这个是不可取的.你需要的这两个功能其实是可以一起实现的,虽然一楼说的很对,当我觉得应该不是你需要的那一种方案.#includemain(){inti=0,k;inta[10;pr
#include <stdio.h>void main (){ int a[10]; int i,k=0,n,min,
#include#includeintmain(void){intn,sum=0,count=0,s;printf("输入一个整数:");scanf("%d",&n);while(n){s
#includeintmat[10][10];voidmain(){intn,i,j;intok=1;scanf("%d",&n);for(i=0;i
#include<stdio.h>#define N 5int main(){ int len(int
charsNum[8];intnumber=0;intnCount=0;cin>>number;charstring[25];memset(string,0,25);itoa(number,strin
//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){\x05intn,sum=0;printf("Typeaninte
#includevoidmain(){\x09inti,n;\x09inta[10];\x09ints,p;\x09printf("n:");\x09scanf("%d",&n);\x09for(i=
x%10=a1,x/10%10=a2,x/100=a3,a=a1*100+a2*10+a3再答:判断是几位:if(a>100)3if(a>1000)4if(a>10000)5再问:再问:能帮我看一下错
#includeintmain(void){intn=0,m=0,i=0,j=0,k=0;scanf("%d",&n);while(n--){\x09scanf("%d",&m);\x09for(j=
#includevoidmain(){inti,n;printf("pleaseinputn:");scanf("%d",&n);printf("%d=",n);for(i=2;i
#include#defineMAXN20intmain(){intn,t,k=0;intia[MAXN];printf("请输入一个整数:");scanf("%d",&n);while(t=n%10
作为字符串形式读入比用整形读入好处理,而且数的位数不受限制(整型有数值超界问题)#includevoidmain(){chars[80];//最长80位数字inti,L;printf("pleasei