PAT 1060. Are They Equal (25)

最新推荐文章于 2022-03-05 17:07:08 发布

小5555

最新推荐文章于 2022-03-05 17:07:08 发布

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本文链接：https://blog.youkuaiyun.com/xtzmm1215/article/details/38809629

本文介绍了一种算法，用于判断两个浮点数在一个只保留指定数量有效数字的机器上是否被视为相等。通过简单的截断而非四舍五入来处理浮点数，并提供了具体的实现代码。

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http://pat.zju.edu.cn/contests/pat-a-practise/1060

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

下面的代码为嘛过不了呢=，=

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
struct ans{
	string d;
	int k;
};
ans convert(string s, int n){
	ans tmp;
	if ((s[0] != '0') && (s.find(".") != string::npos)){
		tmp.k = s.find(".");
		if (s.size() - 1 >= n){
			if (tmp.k >= n)
				tmp.d = s.substr(0, n);
			else if (tmp.k < n){
				tmp.d = s.substr(0, tmp.k) + s.substr(tmp.k+1, n-tmp.k);
			}
		}
		else{
			tmp.d = s.substr(0, tmp.k) + s.substr(tmp.k + 1, s.size()-1-tmp.k);
			for (int i = 0; i < n - (s.size() - 1); i++){
				tmp.d += "0";
			}
		}
	}
	if ((s[0] != '0') && (s.find(".") == string::npos)){
		tmp.k = s.size();
		if (n <= tmp.k)
			tmp.d = s.substr(0, n);
		else if (n > tmp.k){
			tmp.d = s;
			for (int i = 0; i < n - tmp.k; i++){
				tmp.d += "0";
			}
		}
	}
	if (s[0] = '0' && s != "0"){
		int t;
		for (int i = 2; i < s.size(); i++){
			if (s[i] != '0'){
				t = i;
				tmp.k = -(i - 1);
				break;
			}
		}
		if (s.size() - t >= n)
			tmp.d = s.substr(t, n);
		else
		{
			tmp.d = s.substr(t, n);
			for (int i = 0; i < n - s.size() + t; i++){
				tmp.d += "0";
			}
		}
	}
	if (s == "0"){
		tmp.k = 0;
		tmp.d = "";
		for (int i = 0; i < n; i++){
			tmp.d += "0";
		}
	}
	return tmp;
}
int main(){
	int n;
	string s1, s2;
	ans tmp1, tmp2;
	cin >> n >> s1 >> s2;
	tmp1 = convert(s1, n);
	tmp2 = convert(s2, n);
	if (tmp1.d == tmp2.d && tmp1.k == tmp2.k)
		cout << "YES" << " " << "0." << tmp1.d << "*10^" << tmp1.k << endl;
	else
		cout << "NO" << " " << "0." << tmp1.d << "*10^" << tmp1.k << " " << "0." << tmp2.d << "*10^" << tmp2.k << endl;
	return 0;
}

我只能说坑太多，各种格式的数字也没描述清楚。。。真操蛋

AC代码：

#include <stdio.h>
#include <string.h>
#define MAX 110
struct result{
	char d[MAX];
	int k;
};

result getResult(char *a, int n){
	result r;
	int firstPos = -1;
	int pointPos = -1;
	int index = 0;
	int i;
	for (i = 0; a[i]; i++){
		if (a[i] == '.'){
			pointPos = i;
			continue;
		}
		else if (a[i] == '0' && firstPos == -1)
			continue;
		else{
			if (firstPos == -1)
				firstPos = i;
			if (index < n)
			{
				if (index < strlen(a))
					r.d[index++] = a[i];
				else
					r.d[index++] = '0';
			}
		}
	}
	r.d[index] = 0;
	if (pointPos == -1)
		pointPos = i;
	if (pointPos - firstPos < 0)
		r.k = pointPos - firstPos + 1;
	else
		r.k = pointPos - firstPos;
	if (index == 0){
		int i;
		for (i = 0; i != n; i++)
			r.d[i] = '0';
		r.d[i] = 0;
		r.k = 0;
	}
	return r;
}

int main(){
	int n;
	char a[MAX], b[MAX];
	scanf("%d%s%s", &n, a, b);
	result r1 = getResult(a, n);
	result r2 = getResult(b, n);
	if (strcmp(r1.d, r2.d) == 0 && r1.k == r2.k)
		printf("YES 0.%s*10^%d\n", r1.d, r1.k);
	else
		printf("NO 0.%s*10^%d 0.%s*10^%d\n", r1.d, r1.k, r2.d, r2.k);
	return 0;
}