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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 12358.9Sample Output 1:
YES 0.123*10^5Sample Input 2:
3 120 128Sample Output 2:
NO 0.120*10^3 0.128*10^3
#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 0for (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 05.02if (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 12345s1new += "0.";for (int i = 0; i < n; i++) {if (i < s1.length())s1new += s1[i];elses1new += ‘0‘;}s1new += "*10^";int temp = s1.length();if (temp == 100)s1new += "100";else if (temp >= 10) {s1new += temp / 10 + ‘0‘;s1new += temp % 10 + ‘0‘;}elses1new += temp + ‘0‘;}else if (dotPos != 1) {//dot not at the second pos, such like 123.45 12.3s1new += "0.";bool dotFlag = false;for (int i = 0; i < n; i++) {if (i < s1.length()) {if (s1[i] == ‘.‘) {dotFlag = true;continue;}s1new += s1[i];}elses1new += ‘0‘;}if (dotFlag == true) {if (n < s1.length())s1new += s1[n];elses1new += ‘0‘;}s1new += "*10^";int temp = dotPos;if (temp == 100)s1new += "100";else if (temp >= 10) {s1new += temp / 10 + ‘0‘;s1new += temp % 10 + ‘0‘;}elses1new += temp + ‘0‘;}else {if (s1[0] == ‘0‘) {//0.23 0.00023 and so ons1new += "0.";int j = 2, cnt = 0, flag = 0;while (true){if (s1[j] == ‘0‘ && flag == 0)cnt++;elsebreak;j++;}for (int i = j; i < j + n; i++) {if (i < s1.length())s1new += s1[i];elses1new += ‘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‘;}elses1new += temp + ‘0‘;}}else {//2.002 1.354 and so ons1new += "0.";for (int i = 0; i <= n; i++) {if (i == 1)continue;if (i < s1.length())s1new += s1[i];elses1new += ‘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;}
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原文地址：http://www.cnblogs.com/zzandliz/p/5023193.html
