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?0.123×10?5???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?10?100??, 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

?解題思路：

見(jiàn)代碼注釋，摘自《算法筆記》

代碼：

#include <stdio.h> #include <string> #include <iostream> using namespace std;int n; //有效位數(shù) string deal(string s,int& e){ //e被改寫(xiě)了，就是在內(nèi)存中的地址上的數(shù)字在本函數(shù)中不斷被改寫(xiě)，所以在主函數(shù)中讀取的時(shí)候是本函數(shù)修改過(guò)的值，這也是int& e，修改內(nèi)存中e的數(shù)值的用意所在 int k=0; //s的下標(biāo)while(s.length()>0&&s[0]=='0'){ //前導(dǎo)數(shù)是0要去掉，如0000234.23這種情況，不加以判斷可能會(huì)在e指數(shù)的數(shù)值上出錯(cuò) s.erase(s.begin()); } if(s[0]=='.'){ //去完了前導(dǎo)數(shù)為0遇到了小數(shù)點(diǎn)，就意味著該數(shù)小于1 s.erase(s.begin()); //去掉小數(shù)點(diǎn)while(s.length()>0&&s[0]=='0'){//再去前導(dǎo)數(shù)為0的數(shù)，但是由于是在小數(shù)點(diǎn)之后，每去掉一個(gè)前導(dǎo)數(shù)是0，e也就是指數(shù)要減一 s.erase(s.begin());e--; } }//理解為小于1的深度搜索 else{ //如果去掉前導(dǎo)數(shù)后不是小數(shù)點(diǎn)，意味著大于1 while(k<s.length()&&s[k]!='.'){ //尋找小數(shù)點(diǎn)，在尋找過(guò)程中呢，記錄下小數(shù)點(diǎn)的位置和經(jīng)過(guò)的數(shù)位（有利于得到指數(shù)） k++;e++;} if(k<s.length()){ //為什么是k<s.length時(shí)有小數(shù)點(diǎn)，因?yàn)槿绻?k==s.length時(shí)，就是k到最后也沒(méi)找到小數(shù)點(diǎn) s.erase(s.begin()+k); //把小數(shù)點(diǎn)刪掉 ，目的是循環(huán)輸出的時(shí)候不用判斷小數(shù)點(diǎn)了 } //可以理解為大于1的深度搜索 }if(s.length()==0){//如果刪除前導(dǎo)零后，s的長(zhǎng)度是0，也就是這個(gè)數(shù)是0，綜上：這是在處理前導(dǎo)零遇到的三種情況 ，分別是去掉前導(dǎo)零后大于1，小于1和等于0，三種情況 e=0; } int num=0;k=0;string res;while(num<n){//n這里是全局變量 if(k<s.length()) res += s[k++]; //只要還有數(shù)字就加到末尾else res += '0';//前面的數(shù)不夠有效位數(shù)的，補(bǔ)零num++; } return res; }int main(){string s1,s2,s3,s4;cin>>n>>s1>>s2;int e1=0,e2=0;s3=deal(s1,e1);s4=deal(s2,e2);if(s3==s4&&e1==e2){ cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;} else{ cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;} return 0; }

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