PAT 甲级 1060 Are They Equal
1060. Are They Equal (25)
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.h> #include <stdlib.h> #include <stdio.h> #include <algorithm> #include <math.h> #include <string> using namespace std; pair<string,int> m; string a,b; string aa,bb; int n; pair<string,int> fun(string a) { int len=a.length(); int begin,end; int tag1=0,tag2=0; for(int i=0;i<len;i++)//去除前缀的0和后缀的0 { if(!tag1&&(a[i]!='0'||(a[i]=='0'&&a[i+1]=='.'))) begin=i,tag1=1; if(tag2==1&&a[i]!='0') end=i; if(tag2==0) end=i; if(a[i]=='.') tag2=1; } string c=a.substr(begin,end-begin+1); int i; for(i=0;c[i];i++)//找到小数点的位置 if(c[i]=='.') break; int j; for(j=0;c[j];j++)//找到第一个不是0的数字位置,即科学技术法中小数点应该放在哪个位置之前, { if(c[j]!='0'&&c[j]!='.') break; } string d="0.";int num=0; for(int ii=j;c[ii];ii++) { if(c[ii]=='.') continue; d+=c[ii]; num++; if(num>=n) break; } for(int i=0;i<n-num;i++)//补0 d+='0'; int k=i-j;//小数点位置的变化,就是指数 if(k<0) k++; m.first=d; m.second=k; return m; } int main() { cin>>n>>a>>b; pair<string,int> mm1,mm2; mm1=fun(a); mm2=fun(b); if(mm1==mm2) cout<<"YES "<<mm1.first<<"*10^"<<mm1.second<<endl; else cout<<"NO "<<mm1.first<<"*10^"<<mm1.second<<" "<<mm2.first<<"*10^"<<mm2.second<<endl; return 0; }