这道题目让我实在是够蛋疼的,所以就发个CNT来祭奠一下...
Problem Description:
If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*105 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.
Input Specification:
Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.
Output Specification:
For each test case, print in a line "YES" if the two numbers are treated equal, and then the number in the standard form "0.d1...dN*10^k" (d1>0 unless the number is 0); or "NO" if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.
Note: Simple chopping is assumed without rounding.
Sample Input 1:3 12300 12358.9Sample Output 1:
YES 0.123*10^5Sample Input 2:
3 120 128Sample Output 2:
NO 0.120*10^3 0.128*10^3
本题的解题思路是比较简单的,不需要用到复杂的算法。
不过小菜做的过程中因为没考虑周全而浪费了很多时间。
本题解题关键在于浮点数的标准化表示形式的转换。(即题目中给定的0.d1d2d3...dn*10^k)
输入数据主要有一下几种情况需要主要:
- A或B输入存在前导零情况;
- A或B是小于0的小数;(做题过程一直把它忽略了,最后才蛋疼的想起来)
- A或B为0;
- 以上3种情况组合而成。
鉴于代码写得实在令人看不下去,就不贴出来污染大家了。