输入一个正整数输出其反序数,c语言 c语言:任意输入一整数,编程输出其反序数

\u8f93\u5165\u4e00\u4e2a\u6b63\u6574\u6570\u8f93\u51fa\u5176\u53cd\u5e8f\u6570,c\u8bed\u8a00

#include
#include
#include
void
main(){
int
ipnumber,i;
char
string[100];
printf("\u8bf7\u8f93\u5165\u8981\u8fdb\u884c\u53cd\u5e8f\u7684\u6b63\u6574\u6570\uff1a");
scanf("%d",&ipnumber);
itoa(ipnumber,string,10);
i=strlen(string);
printf("\n\u53cd\u5e8f\u540e\u7684\u6b63\u6574\u6570\u4e3a\uff1a");
while(i>=0)
printf("%c",string[--i]);
printf("\n");
}
\u5229\u7528\u51fd\u6570\u5b9e\u73b0\u8f83\u4e3a\u7b80\u5355\uff0c\u4f46\u662f\u8f93\u5165\u7684\u6574\u6570\u5927\u5c0f\u6709\u4e00\u5b9a\u7684\u9650\u5236\uff0c\u8fd9\u91cc\u4e0d\u80fd\u8d85\u8fc710\u4f4d\uff0c\u8d85\u8fc7\u540e\u56e0\u4e3a\u8d85\u8fc7\u4e86\u6574\u5f62\u7684\u8868\u793a\u8303\u56f4\uff0c\u4ece\u800c\u4e0d\u80fd\u5c06\u5176\u6b63\u786e\u7684\u53cd\u5e8f\uff0c\u5982\u679c\u60f3\u4e0d\u53d7\u8868\u793a\u8303\u56f4\u7684\u9650\u5236\uff0c\u53ef\u4ee5\u5bf9\u6574\u6570\u8fdb\u884c\u5206\u5272\uff08\u5229\u7528\u5bf910\u6c42\u4f59\u548c\u5bf910\u8fdb\u884c\u6574\u9664\uff09\uff0c\u518d\u9010\u4f4d\u7684\u8f93\u51fa\u3002

#includemain(){int n,s=0;scanf("%d",&n);while(n){s*=10;s+=n%10;n/=10;}printf("%d",s);}\u60a8\u597d\uff0c\u5f88\u9ad8\u5174\u4e3a\u60a8\u89e3\u7b54\uff0c\u6d41\u661f\u4e0d\u6b7b\u4e3a\u60a8\u7b54\u7591\u89e3\u60d1
\u5982\u679c\u672c\u9898\u8fd8\u6709\u4e0d\u61c2\u7684\u5730\u65b9\u8bf7\u8ffd\u95ee\uff0c\u671b\u91c7\u7eb3\u6211\u7684\u56de\u7b54\u3002
\u795d\u697c\u4e3b\u5b66\u4e60\u8fdb\u6b65\u3002
\u4e0a\u9762\u7684\u4ee3\u7801\u53ef\u4ee5\u76f4\u63a5\u590d\u5236\u5230\u7f16\u8bd1\u5668\u91cc\uff0c\u5e26\u6362\u884c\u3002

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void main(){
int ipnumber,i;
char string[100];
printf("请输入要进行反序的正整数:");
scanf("%d",&ipnumber);
itoa(ipnumber,string,10);
i=strlen(string);
printf("\n反序后的正整数为:");
while(i>=0)
printf("%c",string[--i]);
printf("\n");
}
利用函数实现较为简单,但是输入的整数大小有一定的限制,这里不能超过10位,超过后因为超过了整形的表示范围,从而不能将其正确的反序,如果想不受表示范围的限制,可以对整数进行分割(利用对10求余和对10进行整除),再逐位的输出。

#include <stdio.h>
void main()
{
int num ;
scanf("%d",&num);
printf("Result = %d\n", 0-num);
}

#include <stdio.h>
void main()
{
int n;
scanf("%d",&n);
while(n)
{printf("%d ",n%10);
n=n/10;
}
}

看在lz这么急的份上 ~

#include <stdio.h>
#include <string.h>

int main()
{
int num;
char s[12];
int i, len;
char t;

printf("input a number : ");
scanf("%d", &num);

sprintf(s, "%d", num);
len = strlen(s);

for(i=0; i < strlen(s)/2; i++)
{
t = s[i];
s[i] = s[len -1 -i];
s[len -1 -i] = t;
}
puts(s);
}

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