1060 Are They Equal (25 分)
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 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.
Each input file contains one test case which gives three numbers N, A and B, where N (<) 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 1, and that its total digit number is less than 100.
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.
3 12300 12358.9
YES 0.123*10^5
3 120 128
NO 0.120*10^3 0.128*10^3
精度模拟题，挺有意思的，特别注意0.000这种情况，我就在这组踩了好久的坑。
3 0 0.000
YES 0.000*10^0
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 int n; 5 string s,ss; 6 7 string getsome(string s,int &pos){ 8 while(s.length() > 0 && s[0] == ‘0‘) 9 s.erase(s.begin()); 10 string st; 11 if(s[0] == ‘.‘){ 12 int i= 1; 13 for(; i < s.length(); i++){ 14 if(s[i] == ‘0‘){ 15 pos--; 16 }else{ 17 break; 18 } 19 } 20 21 for(; i < s.length(); i++){ 22 st += s[i]; 23 } 24 if(st.length() == 0){ 25 pos = 0; 26 } 27 28 }else{ 29 bool flag = true; 30 for(int i = 0; i < s.length(); i++){ 31 if(s[i] != ‘.‘){ 32 st += s[i]; 33 if(flag) 34 pos++; 35 }else{ 36 flag = false; 37 } 38 } 39 } 40 while(st.length() > n){ 41 st = st.substr(0,n); 42 } 43 while(st.length() < n){ 44 st += "0"; 45 } 46 return st; 47 } 48 49 50 int main(){ 51 cin >> n >> s >> ss; 52 int pos1=0,pos2=0; 53 string s1 = getsome(s,pos1); 54 string s2 = getsome(ss, pos2); 55 if(s1==s2 && pos1==pos2){ 56 cout <<"YES 0."<<s1<<"*10^"<<pos1<<endl; 57 }else{ 58 cout <<"NO 0."<<s1<<"*10^"<<pos1<<" 0."<<s2<<"*10^"<<pos2<<endl; 59 } 60 return 0; 61 }
原文：https://www.cnblogs.com/zllwxm123/p/11191401.html