思路:很简单,就是将要转换的那个不断对转化的进制数不断取余,然后倒叙输出即可
代码:
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#define ll long long
char shuzi[] ="0123456789ABCDEF";
int main()
{
int n,m,i,j,k=0;
char s[100];
scanf("%d",&n);
memset(s,0,sizeof(s));
for(i = 2; i <= 16; i++)
{
printf("%d转换成%d进制的结果为:",n,i);
m=n;
while(m!=0)
{
s[k++]=shuzi[m%i];
m/=i;
}
for(j=k; j>=0; j--)
printf("%c",s[j]);
printf("\n");
memset(s,0,sizeof(s));
k=0;
}
return 0;
}