An archeologist seeking proof of the presence of extraterrestrials in the
Earth's past, stumbles upon a partially destroyed wall containing strange
chains of numbers. The left-hand part of these lines of digits is always
intact, but unfortunately the right-hand one is often lost by erosion of the
stone. However, she notices that all the numbers with all its digits intact
are powers of 2, so that the hypothesis that all of them are powers of 2 is
obvious. To reinforce her belief, she selects a list of numbers on which it is
apparent that the number of legible digits is strictly smaller than the number
of lost ones, and asks you to find the smallest power of 2 (if any) whose
first digits coincide with those of the list.
The Archeologists' Dilemma
|The Archeologists' Dilemma|
Thus you must write a program such that given an integer, it determines (if it exists) the smallest exponent E such that the first digits of 2E coincide with the integer (remember that more than half of the digits are missing).
InputIt is a set of lines with a positive integer N not bigger than 2147483648 in each of them.
OutputFor every one of these integers a line containing the smallest positive integer E such that the first digits of 2E are precisely the digits of N, or, if there is no one, the sentence ``no power of 2".
1 2 10
- Let P denotes the prefix, T denotes the # of lost digits. We are searching
- for N, that the prefix of 2^N is P.
- We have an inequlity of
- P*10^T <= 2^N < (P+1)*10^T
- log2(P*10^T) <= log2(2^N) < log2((P+1)*10^T),
- which is
- log2(P)+T*log2(10) <= N < log2(P+1)+T*log2(10).
- Also, we know that
- P < 10^(T-1),
- that is
- T > log10(P)+1.
- Then, we can brute force on T and find the minmum N.
Source Code://Tue Mar 22 18:11:01 CDT 2011
using namespace std;
long double solve(long double number)
long double lower = log2l(number);
long double upper = log2l(number + 1);
long double f = log2l(10);
long double inc = ceill(log10l(number + 0.5)) + 1;
for (; ceill(lower + inc * f) != floorl(upper + inc * f); inc += 1);
return ceill(lower + inc * f);
int main(int argc, char* argv)
//freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
long double str;
while (cin >> str)
cout << (long long) solve(str) << endl;