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2013
11-12

POJ 3191 The Moronic Cowmpouter [解题报告] Java

The Moronic Cowmpouter

问题描述 :

Inexperienced in the digital arts, the cows tried to build a calculating engine (yes, it’s a cowmpouter) using binary numbers (base 2) but instead built one based on base negative 2! They were quite pleased since numbers expressed in base −2 do not have a sign bit.

You know number bases have place values that start at 1 (base to the 0 power) and proceed right-to-left to base^1, base^2, and so on. In base −2, the place values are 1, −2, 4, −8, 16, −32, … (reading from right to left). Thus, counting from 1 goes like this: 1, 110, 111, 100, 101, 11010, 11011, 11000, 11001, and so on.

Eerily, negative numbers are also represented with 1′s and 0′s but no sign. Consider counting from −1 downward: 11, 10, 1101, 1100, 1111, and so on.

Please help the cows convert ordinary decimal integers (range -2,000,000,000..2,000,000,000) to their counterpart representation in base −2.

输入:

Line 1: A single integer to be converted to base −2

输出:

Line 1: A single integer with no leading zeroes that is the input integer converted to base −2. The value 0 is expressed as 0, with exactly one 0.

样例输入:

-13

样例输出:

110111

温馨提示:

Explanation of the sample:

Reading from right-to-left:

1*1 + 1*-2 + 1*4 + 0*-8 +1*16 + 1*-32 = -13

解题代码:

/* @author: */
import java.util.Scanner;
public class Main {
	

 public static void main(String[] args) {
  Scanner sc = new Scanner(System.in);
  int i, n, h;
  boolean s[]=new boolean[100];
  n=sc.nextInt();
  if( n == 0 ) {
	System.out.printf( "0\n" );
	return ;
   }

   i = 0;
   h = 1;
   while( n!=0 ) {
	if( (n & 1)!=0 ) {
         s[i] = true;
	  n -= h;
	}
	else
	  s[i] = false;
	h = -h;
	n >>= 1;
	i++;
    }

    while(( i-- )!=0)
	System.out.printf( "%d", s[i]?1:0 );

    System.out.println();
   }
}

  1. #include <cstdio>
    #include <cstring>

    const int MAXSIZE=256;
    //char store[MAXSIZE];
    char str1[MAXSIZE];
    /*
    void init(char *store) {
    int i;
    store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
    for(i=’F';i<=’Z';++i) store =i-5;
    }
    */
    int main() {
    //freopen("input.txt","r",stdin);
    //init(store);
    char *p;
    while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
    if(p=fgets(str1,MAXSIZE,stdin)) {
    for(;*p;++p) {
    //*p=store[*p]
    if(*p<’A’ || *p>’Z') continue;
    if(*p>’E') *p=*p-5;
    else *p=*p+21;
    }
    printf("%s",str1);
    }
    fgets(str1,MAXSIZE,stdin);
    }
    return 0;
    }

  2. 我没看懂题目
    2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5
    我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
    第二个是7 0 6 -1 1 -6 7输出是14 7 7
    不知道题目例子是怎么得出来的