请问我这个十进制转化二进制程序哪里有问题？
问题描述：
#include
int main ()
{
int n,x,d,m=0;
scanf("%d",&n);
while(n--)
{
scanf("%d",&x);
if(x>0&&x<=10000)
m=0;
d=x;
do
{
m=m*10+x%2;
x=x/2;
}while(x!=0);
if(d%2!=0)
printf("%d\n",m);
if(d%2==0)
printf("%d\n",m*10);
}
return 0;
}
它编译无问题，就是运行时输入2，12等数时出现问题，5，10却没问题
答
#include <stdio.h>
#include <math.h>
int main()
{
int n, x, d, m = 0;
scanf_s("%d", &n);
while (n--)
{
scanf_s("%d", &x);
if (x>0 && x <= 10000)
m = 0;
d = x;
for (int i = 0; x != 0; i++, x = x / 2){
m = (x % 2) * pow((double)10, i - 1) + m;
}
if (d % 2 != 0)
{
if (d == 1)
printf("%d\n", d);
else
printf("%d\n", m);
}
if (d % 2 == 0)
printf("%d\n", m * 10);
}
return 0;
}
答
#include <stdio.h>
#include <math.h>
int main ()
{
int n,x,d,m=0;
scanf("%d",&n);
while(n--)
{
scanf("%d",&x);
if(x>0&&x<=10000)
m=0;
d=x;
for(int i = 0;x != 0; i++,x = x/2 ){
m = (x%2) * pow((double)10, i - 1) + m;
}
if(d%2!=0)
printf("%d\n",m);
if(d%2==0)
printf("%d\n",m*10);
}
return 0;
}