1060. Are They Equal (25)

1060. Are They Equal (25)
时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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 12358.9
Sample Output 1:
YES 0.123*10^5
Sample Input 2:
3 120 128
Sample Output 2:
NO 0.120*10^3 0.128*10^3

注意点：

int pos = str.find(‘.’);
对于正确删除，应用迭代器的方式：
str.erase(str.begin() + pos, str.begin() + pos + 1);//去掉.
str.erase(pos,1);

注意str.erase(pos)会删除pos以及以后的所有字符

#define _CRT_SECURE_NO_WARNINGS
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>

using namespace std;

struct FloatN
{
    string m;
    int e;

    FloatN()
    {
        e = 0;
    }
};

void change(FloatN &N, int b)
{
    int idx = 0;
    string tmp = N.m;
    N.e = 0; N.m.clear();
    N.m = "0.";
    while (tmp.size() && tmp[0] == '0') // 除去前导为0的位
        tmp.erase(tmp.begin());

    if (tmp.size())//要么有小数点，要么没有小数点有整数部分
    {
        if (tmp[0] == '.')
        {
            tmp.erase(tmp.begin());
            while (tmp.size() && tmp[0] == '0')
            {
                --N.e;
                tmp.erase(tmp.begin());
            }
            if (tmp.size())
            {
                for (size_t i = 0; i < tmp.size() && idx < b; ++i, ++idx)
                    N.m += tmp[i];
            }
            else
                N.e = 0;
        }
        else //整数情况
        {
            int pos = tmp.find('.');
            if (pos != string::npos)
            {
                N.e = pos;
                tmp.erase(tmp.begin() + pos, tmp.begin() + pos + 1);//去掉.
            }
            else
            {
                N.e = tmp.size();
            }
            for (size_t i = 0; i < tmp.size() && idx < b;++i, ++idx)
                N.m += tmp[i];

        }
    }

    while (idx < b)  //若有剩下的位为0，则补齐
    {
        N.m += '0';
        ++idx;
    }
}


int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    int n;
    FloatN a, b;
    cin >> n >> a.m >> b.m;
    change(a, n);
    change(b, n);
    if (a.m == b.m && a.e == b.e)
    {
        cout << "YES " << a.m << "*10^" << a.e;
    }
    else
    {
        cout << "NO " << a.m << "*10^" << a.e << " " << b.m << "*10^" << b.e;
    }

    return 0;
}