PAT甲级 Are They Equal (25) （恶心模拟）-蒲公英云

PAT甲级 Are They Equal (25) （恶心模拟）

男娘i 2021-12-03 11:04 291阅读 0赞

Are They Equal (25)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB

题目描述

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.

输入描述:

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.

输出描述:

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.

输入例子:

3 12300 12358.9

输出例子:

YES 0.123*10^5
题解：
　　这个题关键不在于各种分情况讨论，而是如何找到有效数位
1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int n;
 7 string s1,s2;
 8 struct node{
 9     string ss;
10     int kk;
11 }e;
12 node judge(string s)
13 {
14     int flag=0;
15     int len=s.length();
16     int k=0;
17     string ans="";
18     for(int i=0;i<len;i++)
19     {
20         if(s[i]=='.')
21             flag=1;
22         else
23         {
24             if(!flag)
25                 k++;
26             if(ans==""&&s[i]=='0')
27             {
28                 k--;
29             }
30             else
31             {
32                 ans+=s[i];
33             }
34         }
35         
36     }
37     while(ans.length()<n)
38         ans+='0';
39     if(ans.length()>n)
40         ans=ans.substr(0,n);
41 //    cout<<"ans="<<ans<<endl;
42     int ok=0;
43     for(int i=0;i<ans.length();i++)
44     {
45         if(ans[i]!='0')
46         {
47             ok=1;
48             break;
49         }
50     }
51     if(!ok)
52         k=0;
53     e.ss=ans;
54     e.kk=k;
55     return e; 
56 }
57 int main()
58 {
59     cin>>n>>s1>>s2;
60     node ans1=judge(s1);
61     node ans2=judge(s2);
62     if(ans1.ss==ans2.ss&&ans1.kk==ans2.kk)
63     {
64         cout<<"YES "<<"0."<<ans1.ss<<"*10^"<<ans1.kk<<endl;
65     }
66     else
67     {
68         cout<<"NO "<<"0."<<ans1.ss<<"*10^"<<ans1.kk<<" "<<"0."<<ans2.ss<<"*10^"<<ans2.kk<<endl;
69     }
70     return 0;
71 }