c语言中怎么倒序输出一个数，无论这个数是几位数，输出的结果都得是四位数，不够在后面添0 C语言程序分析,输入一个数,确定是几位数,输出每一位和按逆序...
C\u8bed\u8a00\u3001\u8f93\u5165\u56db\u4f4d\u6570\u3001\u5012\u5e8f\u8f93\u51fa\u3002\u8fd9\u9053\u9898\u9700\u8981\u7528\u5230\u6574\u6570\u7684\u5206\u89e3\uff0c\u6e90\u4ee3\u7801\u5982\u4e0b\uff1a
#include main(){int num,result,a,b,c,d;printf("\u8bf7\u8f93\u5165\u8981\u5206\u89e3\u7684\u56db\u4f4d\u6b63\u6574\u6570\uff1a\n");scanf("%d",&num);a=num%10;c=num/100%10; b=num/10%10; d=num/1000 ; result=a*1000+b*100+c*10+d; printf("%d\u7684\u9006\u5e8f\u6570\u4e3a%d\n",num,result);}
\u8fd0\u884c\u7ed3\u679c\u5982\u4e0b\uff1a
\u6269\u5c55\u8d44\u6599\u6211\u4eec\u53ef\u4ee5\u901a\u8fc7\u5341\u8fdb\u5236\u7684\u53d6\u6574\u53d6\u4f59\u7279\u70b9\u6765\u5206\u522b\u53d6\u5230\u4e2a\u4f4d\u3001\u5341\u4f4d\u3001\u767e\u4f4d\u7684\u6570\u5b57\uff1a
\u767e\u4f4d\uff1aa/100; \u5bf9a \u8fdb\u884c\u9664\u4ee5100 \u90a3\u4e48\u53ef\u4ee5\u53d6\u5230\u767e\u4f4d\uff0c
\u4e2a\u4f4d\uff1aa%10;\u5bf9a \u8fdb\u884c10 \u7684\u53d6\u4f59\u90a3\u4e48\u4e00\u5b9a\u4f1a\u53d6\u5230\u6700\u540e\u4e00\u4f4d\uff08\u5373\u4e2a\u4f4d\uff09\uff0c
\u5341\u4f4d\uff1a a/10%10\uff1b\u6700\u6700\u590d\u6742\u7684\u5c31\u662f\u53d6\u5341\u4f4d \u7684\u6570\u5b57 \u5176\u5b9e\u7531\u53d6\u4e00\u5934\uff08\u767e\u4f4d\uff09\u4e00\u5c3e\uff08\u5341\u4f4d\uff09\u53ef\u4ee5\u77e5\u9053 \u6211\u4eec\u53ef\u4ee5\u5c06\u4e09\u4f4d\u6570\u53d8\u6210\u4e24\u4f4d\u6570\uff0c\u7136\u540e\u5bf9\u4e24\u4f4d\u6570\u53d6\u4f59\u5c31\u53ef\u4ee5\u53d6\u5230\u5341\u4f4d\u7684\u6570\u5b57\u3002
#include #include char *mFun(char *p, int len) //\u98a0\u5012\u6570\u5b57\u7684\u51fd\u6570{ char sp; for(int i=0;i<len/2;i++) { sp = p[i]; p[i] = p[len-i-1]; p[len-i-1] = sp; } return p;}char* qling(char *p, int len) //\u4ece\u9996\u4f4d\u67e5\u627e\u975e0\u5b57\u7b26,\u5e76\u4e14\u8fd4\u56de\u975e0\u9996\u5730\u5740{ if(p[0] == '0')//\u5982\u679c\u8f93\u5165\u7684\u5b57\u7b26\u4e32\u9996\u4f4d\u6ca1\u67090\u5219\u4e0d\u8fd0\u884c\u4e0b\u9762\u7684\u7b97\u6cd5\u76f4\u63a5\u8df3\u8fc7 { for(int i=0;i<len;i++) { if((p[i] == '0') && (p[i+1] != '0')) { return &p[i+1]; } } } return 0;}int main(int argc, char* argv[]){ char sbuf[255]={0}; printf("Put Number:"); gets(sbuf); //\u8f93\u5165\u6570\u5b57 int len = strlen(sbuf); //\u8c03\u7528\u51fd\u6570\u7b97\u51fa\u8f93\u5165\u5b57\u7b26\u4e32\u7684\u957f\u5ea6 char *p = qling(sbuf, len); if(p == 0)p = sbuf; //\u5982\u679c\u8fd4\u56de\u4e3a0,\u5219\u8868\u793a\u9996\u4f4d\u5230\u6700\u9ad8\u4f4d\u6ca1\u67090 len = strlen(p); printf("\u4f4d\u6570\u662f%d\u4f4d:%s\n",len,mFun(p,len)); //\u8c03\u7528\u51fd\u6570,\u98a0\u5012\u6570\u5b57\u4f4d\u7f6e,\u8fd4\u56de\u9996\u5730\u5740,\u5e76\u4e14\u8f93\u51fa getchar(); return 0;}
将数字 按字符串处理，然后 倒序输出，不足补0
代码如下
#include <stdio.h>
#include <string.h>
int main()
{
char s[20];
int i,l;
scanf("%s",s);
l = strlen(s);
for(i = l-1; i >=0 ; i --)
printf("%c", s[i]);
for(; l<4; l ++)
printf("0");
return 0;
}
核心代码：
scanf("%d",&n); int i = 1;
while (n>0 && i<=4){
printf("%d",n % 10);
n /= 10;
i++;
}
for (int j=i;j<=4;j++) printf("0");请采纳
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