UVa_00701_The_Archeologists'_Dilemma.cpp

Problem Links:

uva00701,

Problem:

The Archeologists' Dilemma

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.
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 2^E coincide with the integer (remember that more than half of the digits are missing).

Input 

It is a set of lines with a positive integer N not bigger than 2147483648 in each of them.

Output 

For every one of these integers a line containing the smallest positive integer E such that the first digits of 2^E are precisely the digits of N, or, if there is no one, the sentence ``no power of 2".

Sample Input 

1
2
10

Solution:

1. Let P denotes the prefix, T denotes the # of lost digits. We are searching
2.   for N, that the prefix of 2^N is P. 
3.   We have an inequlity of
4.     P*10^T <= 2^N < (P+1)*10^T
5.   thus
6.     log2(P*10^T) <= log2(2^N) < log2((P+1)*10^T),
7.   which is
8.     log2(P)+T*log2(10) <= N < log2(P+1)+T*log2(10).
9.   Also, we know that
10.     P < 10^(T-1),
11.   that is
12.     T > log10(P)+1.
13.   Then, we can brute force on T and find the minmum N.

Source Code:

//Tue Mar 22 18:11:01 CDT 2011
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

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;
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}