HDU 2031 进制转换

进制转换

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58687    Accepted Submission(s): 31957

Problem Description

输入一个十进制数N，将它转换成R进制数输出。

 

Input

输入数据包含多个测试实例，每个测试实例包含两个整数N(32位整数)和R（2<=R<=16, R<>10）。

 

Output

为每个测试实例输出转换后的数，每个输出占一行。如果R大于10，则对应的数字规则参考16进制（比如，10用A表示，等等）。

 

Sample Input

7 2 23 12 -4 3

 

Sample Output

111 1B -11

 

Author

lcy

 

Source

C语言程序设计练习（五）

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<map>
#include<sstream>
#include<queue>
#include<stack>
#include<cmath>
#include<list>
#include<vector>
#include<string>
using namespace std;
#define long long ll
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int N = 500005;
int n, m, tot;
int a[N];
stack<char> s;
int main()
{
    while(cin >> n >> m )
    {
        if(n<0)
            cout<<"-";
        int tmp;
        while(n)
        {
            tmp = n%m;
            tmp = tmp>0?tmp:-tmp;
            if(tmp<10) s.push(tmp +'0');
            else s.push(tmp - 10 + 'A');
            n /= m;
        }
        while(!s.empty())
        {
            cout<<s.top();
            s.pop();
        }
        cout<<endl;
    }
}

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<map>
#include<sstream>
#include<queue>
#include<stack>
#include<cmath>
#include<list>
#include<vector>
#include<string>
using namespace std;
#define long long ll
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int N = 500005;
int n, m, tot;
int a[N];
stack<char> s;
void print(int a)
{
    if(a<10) cout<<a;
    else cout<<(char)(a-10+'A');
}
int main()
{
    while(cin >> n >> m )
    {
        if(n<0)
            cout<<"-",n=-n;
        int tmp, k = 0;
        while(n)
        {
            a[++k] = n % m;
            n /= m;
        }
        for(int i=k; i>=1; i--)
        {
            print(a[i]);
        }
        cout<<endl;
    }
}