c语言编程,输入一个十进制数,转换为二,八,十六进制输出,用数组。 两个字符串连接成一个字符串。拜 C语言编程:从键盘上输入一个十进制数,将其转换成八进制数,然...

\u5173\u4e8ec\u8bed\u8a00\u7684\u95ee\u9898\uff0c\u8f93\u51fa\u6570\u7ec4s\u4e2d\u5b57\u7b26\u4e32\uff0c\u5b50\u4e32\u4eces[begin]\u5f00\u59cb\uff0c\u957f\u5ea6\u4e3alen\u3002

int location;

for (location = begin; location <= begin + len; location++)
printf("%c", s[location]);

\u9898\u76ee\u7684\u610f\u601d\u597d\u50cf\u662f\u5728\u4e0b\u9762\u6dfb\u52a0\u4ee3\u7801- -\uff1a
for (; len > 0; len--, begin++)
printf("%c", s[begin]);

\u975e\u9012\u5f52\u7b97\u6cd5\uff1a
#include
using namespace std;
#include
#define MAXSIZE 20
typedef struct Stack
{
char node[MAXSIZE];
int top;
}Stack;

int main()
{


Stack s;
s.top=-1;
int n;
cout<<"\u8bf7\u8f93\u5165\u5341\u8fdb\u5236\u6570\uff1a"<<endl;
cin>>n;
int m;
cout<<"\u8bf7\u8f93\u5165\u8981\u8f6c\u5316\u7684\u8fdb\u5236\u6570\uff1a"<<endl;
cin>>m;
int q=n;
while(n!=0)
{
q=n%m;
n=n/m;
s.top++;
if(q<10)
s.node[s.top]=q+48;
else
s.node[s.top]=(q-9)+64;
}
while(s.top>=0)
{
cout<<s.node[s.top--];
}

return 0;
}

\u9012\u5f52\u7b97\u6cd5\uff1a\uff08\u672a\u91c7\u7528\u6570\u7ec4\u5b58\u50a8\uff09
#include
using namespace std;

#define N 8
void print(int M);
int main()
{
print(100);

return 0;
}
void print(int M)
{
int n=M%N;
M=M/N;
if(M!=0)
print(M);
cout<<n;
}

你好,参考如下代码:

#include<iostream>
using namespace std;
#include "string.h"
#include <stdlib.h>
#include <stdio.h>

//10-15数字转换为字母
char i2a(int val)
{
    if(val<10 || val>15)
        return '0'+ val;
    const char* a = {"ABCDEF"};
    return a[val-10];
}
int main()
{
    while(1)
{
    cout<<"输入待转换数字(输入0或负数结束):"<<endl;
    int a;
    cin>>a;
    if(a<=0)
        return 0;


    cout<<"输入基数(2-16之间):"<<endl;
    int b;
    cin>>b;
    if(b<2 || b>16) 
    {
        cout<<"输入基数错误!"<<endl;
        return 0;
    }

    char c[100];
    int i=0;    
    for(;;)
    {
        if(a<b)
        {
            c[i++] = i2a(a);
            break;
        }
        else
        {
            
            while(a>=b)
            {
                int d = a%b;

                c[i++] = i2a(d);

                a = a/b;
            }
            c[i++] = i2a(a);
            break;
        }
    }

    char d[100];
    cout<<"转换后的数字:";
    for(int j=0;j<i;j++)
    {
        d[j] = c[i-j-1];
        cout<<d[j];
    }
    cout<<endl;
    
    //cout<<"转换后的数字:"<<d<<endl;
}
}


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