进制转换

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 23992 Accepted Submission(s): 13432

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 

#include "stdio.h"#include #include 
         #include 
          using namespace std;int main(){    int num , decimal,chushu,yushu,i;    map
            bitmap;    char ch;    bool single = false;    bool IsZheng;    while(scanf("%d %d",&num,&decimal) != EOF)    {        bitmap.clear();        if(num < 0)        {            num = 0 - num;            IsZheng = false;        }else        {            IsZheng = true;        }        i = 0;        chushu = num / decimal;        single = false;        while(1)        {            yushu = num % decimal;            if(yushu > 9)            {                ch = yushu - 10 + 65;                bitmap.insert(pair
            (i,ch));            }else            {                ch = yushu + 48;                bitmap.insert(pair
             (i,ch)); } num = chushu; if(single == true) { break; } chushu = num / decimal; if(chushu == 0) { single = true; } i++; } i = 0; if(!IsZheng) { cout << "-"; } for (map
              ::reverse_iterator num = bitmap.rbegin(); num != bitmap.rend(); num++) { if (i != (int)bitmap.size()-1) { cout << num->second; } else { cout << num->second << endl; } i++; } }}
 