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 0.123×10?5??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 10?100??, 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.d[1]...d[N]*10^k (d[1]>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.
解答：
天将降大任于斯人也，必先苦其心志，劳其筋骨。。
不说了，今天上午都花在了这道题上，我的代码始终23分，有一个用例不过，下午借鉴他人的实现思路，终于AC!
思路:
（1）用两个值firstPos和pointPos保存一个浮点数的第一个非零值位置和小数点位置，这样可以完美的解决0023.34、0.0014等用例测试；
（2）用一个字符串保存基数，第一个位置就是firstPos所指的位置；
（3）对（1）中的情况进行补充，如果出现整数，即123，那么pointPos就被初始化为s.size()，即默认在最后面添加了小数点；如果出现数0，那么对firstPos和pointPos同时初始化为s.size();
（4）输出时，判定基数是否相同，同时指数是否相同。
AC代码如下：
#include<iostream>
#include<string>
#include<cstdio>using namespace std;typedef struct{string s; //存储0.xxxxxx的xxxxxx部分 int bit; //存储指数
}num;int n;
string a, b;num handle(const string& a)
{num tmp;int firstPos = -1, pointPos = -1;for(int i = 0; i < a.size(); ++i){if(a[i] == '.'){pointPos = i; } else if(isdigit(a[i]) && a[i] != '0' && firstPos == -1){firstPos = i;tmp.s.push_back(a[i]);}else if(firstPos != -1 && isdigit(a[i])){tmp.s.push_back(a[i]);}}//如果没有小数点，那么默认在整数最后面有小数点 if(pointPos == -1) pointPos = a.size();//firstPos是-1,那么数字一定是0 if(firstPos == -1){firstPos = pointPos = a.size();} //对数进行补0 if(tmp.s.size() < n){tmp.s.append(n - tmp.s.size(), '0');}else{tmp.s = tmp.s.erase(n);}//cout << pointPos << " " << firstPos << endl; if(pointPos < firstPos) tmp.bit = pointPos - firstPos + 1;else tmp.bit = pointPos - firstPos;return tmp;
}
int main()
{cin >> n >> a >> b;num result1, result2;result1 = handle(a);result2 = handle(b); if(result1.s == result2.s && result1.bit == result2.bit){cout << "YES 0." << result1.s << "*10^" << result1.bit << endl;}else {cout << "NO";cout << " 0." << result1.s << "*10^" << result1.bit;cout << " 0." << result2.s << "*10^" << result2.bit << endl;}return 0;
}