A1060 Are They Equal (25 分)

代码长度限制 16 KB
时间限制 400 ms
内存限制 64 MB

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.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.

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

Coding:

#include <iostream>
#include <cmath>
#include <string>
using namespace std;

int main() {
    int N;
    string A, B, a, b;
    cin >> N >> A >> B;
    int lena = A.size(), lenb = B.size();
    int dna = lena, dnb = lenb;

    for (int i = 0; i < lena; i++) {
        if (A[i] == '.') {
            dna = i;
            break;
        }
    }
    for (int i = 0; i < lenb; i++) {
        if (B[i] == '.') {
            dnb = i;
            break;
        }
    }

    int p = 0, q = 0;
    while (A[p] == '0' || A[p] == '.') p++;
    while (B[q] == '0' || B[q] == '.') q++;
    int ea = 0, eb = 0;
    if (p < dna)
        ea = dna - p;
    else
        ea = dna - p + 1;

    if (q < dnb)
        eb = dnb - q;
    else
        eb = dnb - q + 1;

    if (p == lena)
        ea = 0;
    if (q == lenb)
        eb = 0;

    for (int i = 0; i < N; ++i) {
        if (p < lena && A[p] != '.')
            a += A[p];
        else if (p >= lena)
            a += "0";
        else if (A[p] == '.')
            i--;
        p++;
    }

    for (int i = 0; i < N; ++i) {
        if (q < lenb && B[q] != '.')
            b += B[q];
        else if (q >= lenb)
            b += "0";
        else if (B[q] == '.')
            i--;
        q++;
    }
    if (a == b && ea == eb)
        cout << "YES 0." << a << "*10^" << ea << endl;
    else
        cout << "NO 0." << a << "*10^" << ea << " " << "0." << b << "*10^" << eb << endl;
    return 0;
}

Eighty Percent