c程序输入01234输出4位数,各个数之和为10
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#include<stdio.h>voidmain(){ inta,m; for(inti=0;
给你个思路,先把输入的数字赋给a;设置一个数组b[10];整数i;i=0;do{b[i]=a%10;a=a/10;i++;}whlie(a>0)这样的话,如果输入412;数组b的b[0]=2、b[1]
#includemain(){inti=0,s=0,n;scanf("%d",&n);while(n>0){s+=n%10;i++;n/=10;}printf("%d%d\n",s,i);}
定义四个变量intn,max,min,sum;先读一个数据scanf("%d",&n);sum=max=min=n;再循环9次,输入9个数,同时,累加,比较大小for(i=0;imax)max=n;i
#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}
#includeintmain(){longa;printf("请输入一个无位数:");scanf("%ld",&a);if(a/10000==a%10&&a/1000%10==a%100/1
int类型所能容纳的数字位数不能超过10.我写的这个程序稍微长了点,但不受int类型容量的限制,能够处理很长的整数输入(由buffer数组的大小决定).这程序只处理纯整数输入.有疑问尽管问. 
给你写了个,运行通过,你看看吧,记得采纳哦O(∩_∩)O~#includeintmain(){\x09intnum,i=0;\x09printf("pleaseinputanumble:");\x09
#includeintmain(){inta,b;scanf("%d%d",&a,&b);printf("%d",a-b);return0;}
#includevoidmain(){inti,j,k,m,n;printf("inputanumber:\n");scanf("%d",&n);i=n/1000;j=(n%1000)/100;k=(
scanf("%d,%d,%d,&a,&b,&c");改成scanf("%d,%d,%d“,&a,&b,&c);printf("thelargerstnumberis%d\n,max");改成prin
#includevoidmain(){inta[9][9],i,j,n;scanf("%d",&n);for(i=0;i
#includevoidmain(){floata,b,c,d,t;printf("Pleaseinputfourinteger:");scanf("%f,%f,%f,%f",&a,&b,&c,&d)
#includeusingnamespacestd;intreve_int(intn){intt=0;while(n){t*=10;t+=n%10;n/=10;}returnt;}voidmain()
#include<stdio.h>int main(){int n,s=0;scanf("%d",&n);while(n){s=
#include#includeusingnamespacestd;intmain(){doubler=1.5,h=3.0,v;coutr;couth;v=3.141592653589*r*r*h;c
#include "stdio.h"int main(){ double x,sum=0,ave; int
#includeintmain(){intn=0;inti=0;intd[4]={0};printf("Inputanum(-99999999):");scanf("%5d",&n);n%=10000
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