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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
思路：本题思路就是去除前导0，而且在使用 str.erase()的时候，里面内置的参数必须是地址，切记。
1 #include <iostream> 2 #include <string> 3 using namespace std; 4 int n; 5 string Change(string str,int &e) 6 { 7 int k=0; 8 while(str.length()>0&&str[0]==‘0‘) 9 { 10 str.erase(str.begin()); 11 } 12 if(str[0]==‘.‘) 13 { 14 str.erase(str.begin()); 15 while(str.length()>0&&str[0]==‘0‘) 16 { 17 str.erase(str.begin()); 18 e--; 19 } 20 } 21 else 22 { 23 while(k<str.length()) 24 { 25 if(str[k]!=‘.‘) 26 { 27 k++; 28 e++; 29 } 30 else 31 { 32 str.erase(str.begin()+k); 33 break; 34 } 35 } 36 } 37 if(str.length()==0) 38 { 39 e=0; 40 } 41 string ans; 42 int i; 43 for(i=0;i<n;i++) 44 { 45 if(i<str.length()) 46 { 47 ans+=str[i];//string需要用+相加 48 } 49 else 50 ans+=‘0‘; 51 } 52 return ans; 53 } 54 int main(int argc, char *argv[]) 55 { 56 string str1,str2; 57 cin>>n>>str1>>str2; 58 int e1=0,e2=0; 59 string ans1=Change(str1,e1); 60 string ans2=Change(str2,e2); 61 if(ans1==ans2&&e1==e2) 62 { 63 cout<<"YES 0."<<ans1<<"*10^"<<e1<<endl; 64 } 65 else 66 { 67 cout<<"NO 0."<<ans1<<"*10^"<<e1<<" 0."<<ans2<<"*10^"<<e2<<endl; 68 } 69 return 0; 70 }
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原文地址：http://www.cnblogs.com/GoFly/p/4294218.html
