A1060. Are They Equal

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 12358.9

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

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;  // 0000.0012   0123.34
 5 char str1[200], str2[200];
 6 int process(char s[], int len){
 7     int i,j, exp;
 8     for(i = 0; s[i] != '\0' && s[i] == '0'; i++);  //去除前导0
 9     if(s[i] != '\0'){
10         for(j = 0; s[j + i] != '\0'; j++)
11             s[j] = s[j + i];
12         s[j] = '\0';
13     }else{
14         for(int k = 0; k < len; k++)
15             s[k] = '0';
16         s[len] = '\0';
17         return 0;
18     }
19     if(s[0] == '.'){  //.0012
20         for(i = 1; s[i] != '\0' && s[i] == '0'; i++);
21         if(s[i] == '\0'){
22             exp = 0;
23             s[0] = '0';
24         }else{
25             exp = i - 1;
26             exp *= -1;
27             for(j = 0; s[i + j] != '\0'; j++)
28                 s[j] = s[j + i];
29             s[j] = '\0';
30         }
31     }else{  //123.456
32         for(i = 0; s[i] != '\0' && s[i] != '.'; i++);
33         if(s[i] == '\0'){
34             exp = i;
35         }else{
36             exp = i;
37             for(; s[i + 1] != '\0'; i++)
38                 s[i] = s[i + 1];
39             s[i] = '\0';
40         }
41     }
42     for(j = 0; s[j] != '\0'; j++);
43     while(j < len){
44         s[j++] = '0';
45     }
46     s[len] = '\0';
47     return exp;
48 }
49 int main(){
50     int N, exp1, exp2;
51     scanf("%d %s %s", &N, str1, str2);
52     exp1 = process(str1, N);
53     exp2 = process(str2, N);
54     int tag = 1;
55     for(int i = 0; str1[i] != '\0'; i++)
56         if(str1[i] != str2[i]){
57             tag = 0;
58             break;
59         }
60     if(tag == 1 && exp1 == exp2){
61         printf("YES 0.%s*10^%d", str1, exp1);
62     }else{
63         printf("NO 0.%s*10^%d 0.%s*10^%d", str1, exp1, str2, exp2);
64     }
65     cin >> N;
66     return 0;
67 }
View Code

总结：

1、题意：将给出的两个数字变成保留指定位小数的科学计数法的数字，之后看其是否相等。

2、主要分为两种数字，大于1和小于1（0.000123， 1234.5678），其次要注意有0000123的情况出现，要先去除前导0.

3、主要要做的就是想办法求出不带小数点的且符合要求的底数。

4、测试样例： 3 0.0 0   

      输出：YES 0.000*10^0

posted @ 2018-02-05 20:47  ZHUQW  阅读(149)  评论(0)  编辑  收藏  举报