PAT (Advanced Level) Practise 1060 Are They Equal (25)

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本文介绍了一种算法，用于判断两个浮点数在特定机器精度下是否等价。通过科学记数法表示并比较两个非负浮点数的有效位数来解决此问题。

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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

题意：给你两个数，将它们用小数点后k位的科学计数法表示，比较一下两个是否一样

解题思路：这题会有前导零的数据，所以关键在于找到第一个有效数字的位置和小数点的位置，然后就可以处理出一个数字小数点后几位的科学计数法的表示方式

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,x1,x2;
char s1[200], s2[200], ans1[200], ans2[200];

void check(char ch1[], char ch2[],int &x)
{
	int len = strlen(ch1);
	int k = len, kk;
	for(int i=0;i<len;i++)
		if (ch1[i] == '.') { k = i; break; }
	kk = -1;
	for(int i=0;i<k;i++)
		if (ch1[i] != '0') { kk = i; break; }
	if (kk == -1)
	{
		int p = -1;
		for(int i=k+1;i<len;i++)
			if (ch1[i] != '0') { p = i; break; }
		if (p == -1)
		{
			ch2[0] = '0', ch2[1] = '.';
			for (int i = 1; i <= n; i++) ch2[i + 1] = '0';
			ch2[n + 2] = '\0';
			x = 0;
		}
		else
		{
			x = -(p - k - 1);
			ch2[0] = '0', ch2[1] = '.';
			int cnt = 2;
			while (p < len&&cnt < n + 2) ch2[cnt++] = ch1[p], p++;
			while (cnt < n + 2) ch2[cnt++] = '0';
			ch2[cnt] = '\0';
		}
	}
	else
	{
		x = k-kk;
		ch2[0] = '0', ch2[1] = '.';
		int cnt = 2;
		while (ch1[kk] != '.'&&kk < len&&cnt < n + 2) ch2[cnt++] = ch1[kk],kk++;
		kk++;
		while(kk<len&&cnt<n+2) ch2[cnt++] = ch1[kk], kk++;
		while (cnt < n + 2) ch2[cnt++] = '0';
		ch2[cnt] = '\0';
	}
}

int main()
{
	while (~scanf("%d%s%s", &n,s1,s2))
	{
		check(s1, ans1,x1);
		check(s2, ans2,x2);
		if (!strcmp(ans1, ans2) && x1 == x2) printf("YES %s*10^%d\n", ans1, x1);
		else printf("NO %s*10^%d %s*10^%d\n", ans1, x1,ans2,x2);
	}
	return 0;
}