1060. Are They Equal (25)

题目连接：https://www.patest.cn/contests/pat-a-practise/1060

原题如下：

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
失落……考察科学计数法，关键在于小数点和第一个有效位的位置的确定。参考了他人的代码http://blog.csdn.net/xtzmm1215/article/details/38809629

 1 #include<stdio.h>
 2 #define MAXN 105
 3 #include<string.h>
 4 #include<stdlib.h>
 5 struct result{
 6     char d[MAXN];
 7     int k;
 8 };
 9 
10 result getResult(char *a,int n)
11 {
12     int firstPos=-1,pointPos=-1;
13     int index=0;
14     result r;
15     int i=0;
16     for ( i=0;a[i];i++)
17     {
18         if (a[i]=='.')
19         {
20             pointPos=i;
21             continue;
22         }
23         if (firstPos==-1 && a[i]=='0')continue;
24         else
25         {
26             if (firstPos==-1 && a[i]!='0')
27             {
28                 firstPos=i;
29             }
30             if (index<n)
31             {
32                 r.d[index++]=a[i];
33             }
34         }
35     }
36         r.d[index]=0;
37 
38         if (pointPos==-1)pointPos=i;
39         if (pointPos-firstPos<0)r.k=pointPos-firstPos+1;
40         else r.k=pointPos-firstPos;
41 
42         int j;
43         if (index==0)
44         {
45             for (j=0;j<n;j++)r.d[j]='0';
46             r.d[j]=0;
47             r.k=0;
48         }
49     return r;
50 }
51 
52 int main()
53 {
54     int N;
55     scanf("%d",&N);
56     char s1[MAXN],s2[MAXN];
57     result r1,r2;
58     scanf("%s %s",s1,s2);
59     r1=getResult(s1,N);
60     r2=getResult(s2,N);
61     if (strcmp(r1.d,r2.d)==0 && r1.k==r2.k)
62     {
63         printf("YES 0.%s*10^%d",r1.d,r1.k);
64     }
65     else printf("NO 0.%s*10^%d 0.%s*10^%d",r1.d,r1.k,r2.d,r2.k);
66     return 0;
67 }

 

来源：https://www.cnblogs.com/wuxiaotianC/p/6439399.html