UVA 701 - The Archeologists' Dilemma(数论)

最新推荐文章于 2024-02-29 21:35:46 发布

lab104_yifan

最新推荐文章于 2024-02-29 21:35:46 发布

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一位寻找外星人在地球历史上存在证据的考古学家发现了一堵刻有奇怪数字序列的残破墙壁。尽管部分数字因石头侵蚀而丢失，但她注意到所有完整的数字都是2的幂次。为验证这一假设，她挑选了一些数字，这些数字的可读部分严格少于缺失的部分，并请求编写一个程序找出最小的2的幂次，其开始的数字与所选列表匹配。

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

1
2
10

Sample Output

7
8
20

题目大意：给出x，求一个e，使得x * 10 ^ y ≤ 2 ^ e < (x + 1) * 10 ^ y。 y要大于x的位数。

思路: 同取log2。去枚举，

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

char X[15];
double x, y;

int main() {
    while (gets(X)) {
	sscanf(X, "%lf", &x);
	y = strlen(X) + 1;
	while (true) {
	    double f = log2(x) + y * log2(10);
	    double l = log2(x + 1) + y * log2(10);
	    if (ceil(f) - floor(l) <= 10e-9) {
		printf("%.0lf\n", ceil(f));
		break;
	    }
	    y ++;
	}
    }
    return 0;
}