701 - The Archeologists' Dilemma
Time limit: 3.000 seconds
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 exponentE such that the first digits of 2E 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 2E are precisely the digits of N, or, if there is no one, the sentence
``no power of 2".
Sample Input
1 2 10
Sample Output
7 8 20
思路：
黑书上说这题可以用连分数逼近？没想出来...（坑）
完整代码：
/*0.023s*/
#include<cstdio>
#include<cmath>
const double log2_10 = log2(10.0);
int main(void)
{
int n, m, i, tot, lower, upper;
double t1, t2;
while (~scanf("%d", &n))
{
t1 = log2(n), t2 = log2(n + 1);
tot = 0;
for (m = n; m; m /= 10)
tot++;
for (i = tot + 1;; ++i)
{
lower = (int)(t1 + i * log2_10);
upper = (int)(t2 + i * log2_10);
if (lower != upper)
{
printf("%d\n", upper);
break;
}
}
}
return 0;
}