pat 1060

最新推荐文章于 2021-12-08 01:35:03 发布

原创最新推荐文章于 2021-12-08 01:35:03 发布 · 656 阅读

0 ·

CC 4.0 BY-SA版权

1060. Are They Equal (25)

时间限制

50 ms

内存限制

32000 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

//数字为0的时候要输出 0.000*10^0 小数点后有n个0
#include <stdio.h>
#include <string.h>

int n;

int change(char str1[], int num1)
{
	int i, j;
	str1[0] = '0';
	str1[1] = '.';
	int len1 = strlen(str1);

	for(i = 2; i < len1; i++) //小数点所在的位置
		if(str1[i] == '.')
			break;

	for(j = 2; j < len1; j++)
		if(str1[j] >= '1' && str1[j] <= '9')
			break;

	if(i >= j)
		num1 = i - j;
	else
		num1 = i - j + 1;

	for(i = 2; j < len1; j++)
	{
		if(str1[j] != '.')
		{
			str1[i++] = str1[j];
		}
	}

	if(i == 2)
	{
		num1 = 0;
		for(; i <= n+1; i++)
			str1[i] = '0';
	}

	str1[i] = 0;

	str1[2+n] = 0;

	return num1;
}

int main()
{
	int i, j;
	char str1[110], str2[110];
	int num1, num2;

	scanf("%d", &n);
	scanf("%s%s", str1+2, str2+2);

	num1 = change(str1, num1);
	num2 = change(str2, num2);

	if(num1 == num2 && strcmp(str1, str2) == 0)
	{
		printf("YES ");
		printf("%s*10^%d\n", str1, num1);
	}
	else 
	{
		printf("NO ");
		printf("%s*10^%d", str1, num1);
		printf(" %s*10^%d\n", str2, num2);
	}
	return 0;
}