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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 12358.9
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
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <iostream> 4 #include <string.h> 5 6 #include <math.h> 7 #include <algorithm> 8 #include <vector> 9 #include <set> 10 #include <string> 11 12 using namespace std; 13 int n; 14 string deal(string s,int &e) 15 { 16 17 while(s.length()>0&&s[0]==‘0‘) 18 { 19 s.erase(s.begin()); 20 } 21 int k=0; 22 if(s[0]==‘.‘) 23 { 24 s.erase(s.begin()); 25 while(s.length()>0&&s[0]==‘0‘) 26 { 27 s.erase(s.begin()); 28 e--; 29 } 30 }else 31 { 32 while(k<s.length()&&s[k]!=‘.‘) 33 { 34 k++; 35 e++; 36 } 37 if(k<s.length()) 38 { 39 s.erase(s.begin()+k); 40 } 41 } 42 if(s.length()==0)e=0; 43 int num=0; 44 k=0; 45 string res; 46 while(num<n) 47 { 48 if(k<s.length())res+=s[k++]; 49 else res+=‘0‘; 50 num++; 51 } 52 return res; 53 } 54 55 int main(){ 56 string s1,s2,s3,s4; 57 cin>>n>>s1>>s2; 58 int e1=0,e2=0; 59 s3=deal(s1,e1); 60 s4=deal(s2,e2); 61 if(s3==s4&&e1==e2) 62 { 63 cout<<"YES 0."<<s3<<"*10^"<<e1<<endl; 64 65 }else 66 { 67 cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl; 68 } 69 70 return 0; 71 }
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原文地址：http://www.cnblogs.com/ligen/p/4307856.html
