Basically Speaking
Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 2532 Accepted Submission (s): 959
Problem DescriptionThe really Neato Calculator Company, Inc. have recently hired your team to help design their Super Neato Model I Calculator. As a computer scientist you suggested to the company so it would be neato if this new calculator could convert among num BER bases. The company thought this is a stupendous idea and have asked your team to come up with the prototype program for doing BAS E conversion. The project manager of the Super Neato Model I Calculator has informed you and the calculator would have the following NEA To features:
It would have a 7-digit display.
Its buttons would include the capital letters A through F in addition to the digits 0 through 9.
It would support bases 2 through 16.
Inputthe input for your prototype program would consist of one base conversion per line. There'll is three numbers per line. The first number would be being the number in the base converting from. The second number is the base of converting from. The third number is the base of you and converting to. There'll be is one or more blanks surrounding (on either side of) the numbers. There is several lines of input and your program should continue to read until the end of file is reached.
Outputthe output is the converted number as it would appear on the display of the calculator. The number should is right justified in the 7-digit display. If the large to appear on the display and then print "ERROR" (without the quotes) right justified in the DISPLA Y.
Sample Input
1111000 2 101111000 2 162102101 3 102102101 3 12312 4 2 1 a 21234567 10 16 ABCD 16 15
Sample Output
1765 7CA ERROR 11001 12d687 D071
Sourcemid-central USA 1995 What's wrong with the value conversion today? Another, note that the number to be converted may have a--z, so string ... : The idea is to first convert to decimal and then convert to R binary.
#include <iostream> #include <string> #include <math.h>using namespace Std;int A (string & str,int m ) {int X=0,b,t,sum=0;int len=str.size () -1;while ((len+1) >0) {if (str[len]>= ' A ' &&str[len]<= ' Z ') b=str [len]-' A ' +10;elseb=str[len]-' 0 '; Sum+=b*pow (m,x); x++;len--;} return sum;} void ls (int a,int c) {char x[1000];int t=0;while (a) {int m=a%c; if (m>9) x[t++]=m-10+ ' A '; else x[t++]=m+ ' 0 '; A/=c;} if (t>7) cout<< " ERROR" <<endl;else{for (int i=0;i<7-t;i++) cout<< ""; for (int j=t-1;j>= 0;j--) Cout<<x[j];cout<<endl;}} int main () {int b,c;string str;while (cin>>str>>b>>c) ls (A (str,b), c); return 0;}
Hangzhou Electric HDU ACM 1335 basically Speaking
