UVA 701 - The Archeologists' Dilemma(数论)
来源:互联网 发布:dsdt提取 mac 编辑:程序博客网 时间:2024/06/05 18:55
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 2Ecoincide 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
1210
Sample Output
7820
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;}
- UVA 701 - The Archeologists' Dilemma(数论)
- UVa 701 The Archeologists' Dilemma
- UVA 701 - The Archeologists\' Dilemma
- uva 701 - The Archeologists' Dilemma
- UVA 701 The Archeologists' Dilemma
- UVa 701 The Archeologists' Dilemma (数学&枚举)
- H题:UVA 701 The Archeologists' Dilemma
- uva 701——The Archeologists\' Dilemma
- 110503 The Archeologists' Dilemma
- programming-challenges The Archeologists' Dilemma (110503) 题解
- UVa Problem Solution: 701 - The Archaeologist's Dilemma
- UVa Problem 701 The Archeologist’s Dilemma (考古学家的烦恼)
- Inspector's Dilemma UVA
- Inspector's Dilemma UVA
- The Archeologist's Dilemma
- UVa 12118 Inspector's Dilemma
- uva 12118 Inspector's Dilemma
- Uva-12118 Inspector's Dilemma
- Opencv2系列学习笔记2(图像的遍历)
- [总结]视频质量评价技术零基础学习方法
- MFC Tab控件使用方法
- HDU 3242 两个字符串的相加和相减
- 不能将参数 2 从“const char *”转换为“LPCWSTR” vs 2010解决
- UVA 701 - The Archeologists' Dilemma(数论)
- MFC ListBox控件使用方法总结
- Linux性能评测工具之一:gprof篇
- 如何减少代码中的分支语句
- node.js 连接mysql数据库 完美教程
- 碰巧解决
- 基于正则表达式的轻量提示插件--InputNotes文本框输入提示插件
- strust2 AOP学习笔记
- LeetCode Partition List 按值分段链表 系统分析