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<iostream>#include<string>#include<vector>#include<algorithm>#include<string.h>using namespace std;int y;string exc(string s1,int N){ string s;char c1='0'; int dot=-1,number=-1,fir=0,c=0,i; for(i=0;i<s1.size();i++){ if(s1[i]=='.') dot=s1.size()-1-i; if(s1[i]!='.'&&s1[i]!='0'){ if(fir==0){ fir=1; number=s1.size()-1-i; } } } y=number-dot; if(y<0) y++; if(number==-1) y=0; for(i=s1.size()-1-number;c<N;i++){ if(i<s1.size()) if(s1[i]!='.'){ s+=s1[i]; c++; } else{} else{ s+=c1; c++; } } return s;}int main(){ string s1,s2; int N,i,c,l1,l2; cin>>N; cin>>s1>>s2; s1=exc(s1,N); l1=y; s2=exc(s2,N); l2=y; if(s1==s2&&l1==l2){ cout<<"YES "; cout<<"0."<<exc(s1,N)<<"*10^"<<l1; } else{ cout<<"NO "; cout<<"0."<<exc(s1,N)<<"*10^"<<l1; cout<<" 0."<<exc(s2,N)<<"*10^"<<l2; }}感想:1.因为输入有100位所以要用字符串处理2.注意如果输入是0和0.00,要输出YES而且次数为03.要考虑小数,不足的位自动补0
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