传送门:
进制转换
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 60988 Accepted Submission(s): 33224
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
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code:
#include#include #include #include #include void f(int n,int r){ int m; if(n==0) return ; else { f(n/r,r); m=n%r; if(m<10) { printf("%d",m); } else { printf("%c",'A'+m-10); } }}int main(){ int n,r; while(~scanf("%d %d",&n,&r)) { if(n==0) printf("%d",0); else if(n<0) { printf("-"); n=-n; } f(n,r); printf("\n"); } return 0;}