PAT 1060. Are They Equal (25)

1060. Are They Equal (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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...d_N*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

很纠结的一道题。描述很简单，处理起来挺麻烦的。思路就是确定出原始数组第一个有效数字的位置和小数点的位置，从而确定科学计数法的指数。但是输出这个指数时很麻烦，不好转换成数组。机智的我想了个办法，把指数放在数组用不到的位置（x[148]），就很方便输出了。这题还有个坑点是俩数字都是0时，要输出 {0.n个0*10^0}，题目说的不清楚，要自己试，很坑。

#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<math.h>
#include<stack>
#include<vector>
using namespace std;


    char a[150],b[150];
    int n;
    char aa[160],bb[160];
    void geet(char x[],char y[])
    {
        int p1=0;
        int p2=2;
        int p3=0;      //确定小数点位置
        int p4=0;     //确定第一个有效数字位置
        for(;p3<strlen(x);p3++)
            if(x[p3]=='.') break;
        for(;p4<strlen(x);p4++)
           if(x[p4]!='0'&&x[p4]!='.') break;
        for(int i=strlen(x);i<140;i++)         //害怕原数组长度不够，扩长度。
            x[i]='0';                
           x[140]=0;
        p1=p4;
        int bit;
        if(p3<p4) bit=p3-p4+1;    //计算科学计数法的指数
        else bit=p3-p4;
        y[0]='0';
        y[1]='.';
        while(1)
        {   if(x[p1]=='.') p1++;
            y[p2++]=x[p1++];
            if(p2==n+2) break;
        }
        y[p2++]='*';
        y[p2++]='1';
        y[p2++]='0';
        y[p2++]='^';
        y[p2++]=0;
        y[148]=bit;

    }

int main()
{
    cin>>n>>a>>b;
    geet(a,aa);
    geet(b,bb);
    int fllag=0;                                //判断两边是不是都是0
     for(int i=0;aa[i]!='*';i++)
        if(aa[i]!='0'&&aa[i]!='.')
     {
        fllag=1;break;
     }
     for(int i=0;bb[i]!='*';i++)
        if(bb[i]!='0'&&bb[i]!='.')
     {
        fllag=1;break;
     }
      if(fllag==0)
      {
        cout<<"YES 0.";
        for(int i=0;i<n;i++)
         cout<<0;
        cout<<"*10^0";
      }
      else {                      //开始正式判断
    int flag=0;
     if(strcmp(aa,bb)==0&&aa[148]==bb[148]) cout<<"YES ";
     else
     { flag=1;cout<<"NO ";}
     int x=aa[148];
     int y=bb[148];
     cout<<aa<<x;
     if(flag==1) cout<<' '<<bb<<y;
      }

     return 0;
}