1060. Are They Equal (25)

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...d_N*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.

3 12300 12358.9

YES 0.123*10^5

3 120 128

NO 0.120*10^3 0.128*10^3

提交代碼

這道題的坑點不少，開始只想到例子的數字，通過在牛客網上測試，一步步發現了其他不同的case，0，0.0005；000001；

基本思路就是求出兩個字符串，進行比較。

c++的string可以直接通過“+”拼接，同時也可以直接拼接一個字符，可以說是非常方便的。

string的substr()函數，substr(a,b)：a表示初試位置，b表示長度；

以下是代碼

#include <bits/stdc++.h>

using namespace std;

int N;
string A, B;

int main()
{
    cin>>N>>A>>B;
    int ka = 0;
    int na = N;
    int kka = 0;
    for(int i = 0; i < A.length(); i++) {
        if(i != 0 && A[i] == ‘.‘ && A[i-1] == ‘0‘) {
            kka = i-1;
            break;
        }
        if(A[i] != ‘0‘) {
            kka = i;
            break;
        }
    }
    A = A.substr(kka, A.length()-kka);
    if(A[0] == ‘0‘){
        for(int i = 2; i < A.length(); i++) {
            if(A[i] != ‘0‘) {
                ka = 2-i;
                break;
            }
        }
    }
    else {
        for(int i = 0; i < A.length(); i++) {
            if(A[i] == ‘.‘) {
                ka = i;
                break;
            }
        }
        if(ka == 0) {
            ka = A.length();
        }
    }
    string strA = "0.";
    int initA = 0;
    if(ka < 0) {
        initA = 2-ka;
        na += initA;
    }
    for(int i = initA; i < na; i++) {
        //cout<< "i:"<< i<< endl;
        if(i < A.length()) {
            if(A[i] == ‘.‘) {
                na++;
            }
            else {
                strA += A[i];
            }
        }
        else {
            strA += "0";
        }
    }
    strA += "*10^";
    if(ka < 0) {
        strA += "-";
        ka = -ka;
        //cout<< "ka:"<< ka<< endl;
    }
    string sa;
    if(ka == 0) sa = "0";
    while(ka != 0) {
        sa += ka%10+‘0‘;
        ka /= 10;
    }
    for(int i = sa.length()-1; i >= 0; i--) {
        strA += sa[i];
    }
    //strA += ka + ‘0‘;

    int kb = 0;
    int nb = N;
    int kkb = 0;
    for(int i = 0; i < B.length(); i++) {
        if(i != 0 && B[i] == ‘.‘ && B[i-1] == ‘0‘) {
            kkb = i-1;
            break;
        }
        if(B[i] != ‘0‘) {
            kkb = i;
            break;
        }
    }
    B = B.substr(kkb, B.length()-kkb);
    if(B[0] == ‘0‘) {
        for(int i = 2; i < B.length(); i++) {
            if(B[i] != ‘0‘) {
                kb = 2-i;
                break;
            }
        }
    }
    else {
        for(int i = 0; i < B.length(); i++) {
            if(B[i] == ‘.‘) {
                kb = i;
                break;
            }
        }
        if(kb == 0) {
            kb = B.length();
        }
    }
    string strB = "0.";
    int initB = 0;
    if(kb < 0) {
        initB = 2-kb;
        nb += initB;
    }
    for(int i = initB; i < nb; i++) {
        //cout<< "i:"<< i<< endl;
        if(i < B.length()) {
            if(B[i] == ‘.‘) {
                nb++;
            }
            else {
                strB += B[i];
                //cout<< "B:"<< B[i]<< endl;
            }
        }
        else {
            strB += "0";
        }
    }
    strB += "*10^";
    if(kb < 0) {
        strB += "-";
        kb = -kb;
    }
    string sb;
    if(kb == 0) sb = "0";
    while(kb != 0) {
        sb += kb%10+‘0‘;
        kb /= 10;
    }
    for(int i = sb.length()-1; i >= 0; i--) {
        strB += sb[i];
    }
    //strB += kb + ‘0‘;

    if(strA == strB) {
        cout<< "YES "<< strA;
    }
    else {
        cout<< "NO "<< strA<< " "<< strB;
    }
    return 0;
}

PAT-1060. Are They Equal (25)