UVa 11121 Base-2 (number theory &-2 progressive & complementary Ideas)

11121-base-2
Time limit:3.000 seconds
Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_ problem&problem=2062

The creator of the universe works in mysterious ways. But
He uses a base ten counting system and likes round numbers.

Scott Adams

Everyone knows about base-2 (binary) integers and base-10 (decimal) integers, but what about base-2? An integer n written in Base-2 is a sequence of digits (BI), writen right-to-left. Each of the which is either 0 or 1 (no negative digits!), and the following equality must.

n = B0 + B1 ( -2) + b2 ( -2) 2 + b3 (-2) 3 + ...

The cool thing is this every integer (including the negative ones) has a unique base--2 representation with no minus Required. Your task is to find this representation.

Input
The "a" of input gives the number of cases, N (at most 10000). N test Cases follow. Each one was a line containing a decimal integer in the range from-1,000,000,000 to 1,000,000,000.

Output
For each test case, output one line containing ' case #x: ' followed by the same integer and written in base-2 with no leading Zeros.
Sample input Output for sample input

4

1

7

-2

0

Case #1:1

Case #2:11011

Case #3:10

Case #4:0

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Ideas:

One idea is: take what we do in the binary system, just make a small change on the line.

Another kind of thinking is a bit ingenious: if the binary corresponds to that one bit P is 1, then n+=1<< (p+1) and finally n into the normal binary. (faster)

Way 1:

/*0.045s*/
    
#include <cstdio>  
    
int array[40];  
    
int main ()  
{  
    int T, N, I, J, case =0 0;  
    scanf ("%d", &t);  
    while (t--)  
    {  
        scanf ("%d", &n);  
        printf ("Case #%d:", ++case);  
        if (n = = 0)  puts ("0");  
        else
        {  
            i = 0;  
            while (n)  
            {  
                array[i++] = n & 1;  
                n = (n-(n & 1))/-2;  
            }  
            for (j = i-1 J >= 0; j--)  
                printf ("%d", array[j]);  
            Putchar (a);  
        }}}  

Way 2:

/*0.019s*/#include <cstdio> #include <cstdlib> typedef long Long LL;  
    
CONST LL one = 1;  
    int main () {ll n, bit;///very unfortunately needs 1<<32, so long int T, cas = 0, p;  
    scanf ("%d", &t);  
        while (t--) {scanf ("%lld", &n);  
        printf ("Case #%d:", ++cas);  
            if (n = = 0) {puts ("0");  
        Continue  
        ///n is converted to n p = (n > 0 1:0) in the sense of 2,///start n = ABS (n);  
        bit = one << p;  
            while (P <=) {if (bit & N) n + = one << (p + 1);///correction  
            p = 2;  
        bit = one << p; ///to the leading 0 for (bit = one << 31; (bit & N) = = 0;  
        Bit >>= 1); Find binary while (bit) {Putchar (bit & n?)  
            ' 1 ': ' 0 ');  
        Bit >>= 1;  
        }Putchar (10);  
return 0; }