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...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:[/b]

3 12300 12358.9

Sample Output 1:[/b]

YES 0.123*10^5

Sample Input 2:[/b]

3 120 128

Sample Output 2:[/b]

NO 0.120*10^3 0.128*10^3

来源： <http://www.patest.cn/contests/pat-a-practise/1060>

#include<string>

#include <iostream>

using namespace std;

int n;

int SearchDot(string s) { //Search the pos of '.'

for (int i = 0; i < s.length(); i++) {

if (s[i] == '.')

return i;

}

return -1;

}

string ChangeToSci(string s1) {

string s1new;

bool isZero = true;//check whether it is 0

for (int i = 0; i < s1.length(); i++) {

if (s1[i] != '0'&&s1[i] != '.')

isZero = false;

}

if (isZero) {

s1new += "0.";

for (int i = 0; i < n; i++)

s1new += '0';

s1new += "*10^0";

return s1new;//if 0 return

}

bool hengji = false;

string temp;

for (int i = 0; i < s1.length()-1; i++) {//remove 0 in the head such as 05.02

if (s1[i+1]>='0'&&s1[i+1]<='9'&&s1[i] == '0'&&hengji == false) {

continue;

}

else {

hengji = true;

}

temp += s1[i];

}

temp += s1[s1.length() - 1];

s1 = temp;

int dotPos = SearchDot(s1);

if (dotPos == -1) {//no dot in the number ,such like 12345

s1new += "0.";

for (int i = 0; i < n; i++) {

if (i < s1.length())

s1new += s1[i];

else

s1new += '0';

}

s1new += "*10^";

int temp = s1.length();

if (temp == 100)

s1new += "100";

else if (temp >= 10) {

s1new += temp / 10 + '0';

s1new += temp % 10 + '0';

}

else

s1new += temp + '0';

}

else if (dotPos != 1) {//dot not at the second pos, such like 123.45 12.3

s1new += "0.";

bool dotFlag = false;

for (int i = 0; i < n; i++) {

if (i < s1.length()) {

if (s1[i] == '.') {

dotFlag = true;

continue;

}

s1new += s1[i];

}

else

s1new += '0';

}

if (dotFlag == true) {

if (n < s1.length())

s1new += s1[n];

else

s1new += '0';

}

s1new += "*10^";

int temp = dotPos;

if (temp == 100)

s1new += "100";

else if (temp >= 10) {

s1new += temp / 10 + '0';

s1new += temp % 10 + '0';

}

else

s1new += temp + '0';

}

else {

if (s1[0] == '0') {//0.23 0.00023 and so on

s1new += "0.";

int j = 2, cnt = 0, flag = 0;

while (true)

{

if (s1[j] == '0' && flag == 0)

cnt++;

else

break;

j++;

}

for (int i = j; i < j + n; i++) {

if (i < s1.length())

s1new += s1[i];

else

s1new += '0';

}

if (cnt == 0) {

s1new += "*10^0";

}

else {

s1new += "*10^-";

int temp = cnt;

if (temp == 100)

s1new += "100";

else if (temp >= 10) {

s1new += temp / 10 + '0';

s1new += temp % 10 + '0';

}

else

s1new += temp + '0';

}

}

else {//2.002 1.354 and so on

s1new += "0.";

for (int i = 0; i <= n; i++) {

if (i == 1)

continue;

if (i < s1.length())

s1new += s1[i];

else

s1new += '0';

}

s1new += "*10^1";

}

}

return s1new;

}

int main(void) {

string s1, s2;

string s1new, s2new;

cin >> n >> s1 >> s2;

if (n == 0) {

cout << "YES 0";

return 0;

}

string s11, s22;

s11 = ChangeToSci(s1);

s22 = ChangeToSci(s2);

if (s11 == s22) {

cout << "YES " << s11;

}

else {

cout << "NO " << s11 << " " << s22;

}

return 0;

}

[/code]

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