1060. Are They Equal (25)

题目如下:

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...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

这个题目的难点在于要处理的数不一定都是大于1的,还可能出现0.015这样的数字,我之前的代码只能处理大于1的情况,后来参考了小5555的代码,发现可以用两个游标指示小数点和第一个非0数的位置,然后判断二者的位置关系来判断数的值。

首先定义一个结构体,来存储基数和次方:

struct result{
	char d[MAX]; // 0.xxx部分
	int k; // 10的k次方
};

然后设计一个函数,对于输入的数字(存储在char*内)数组进行遍历。

遍历时定义firstPos记录第一个非0数,pointPos记录小数点位置。

需要注意以下问题:

①题目的case似乎出现了00123.45这种坑爹的情况,因此要处理无效的0,也就说开头出现的0是不能要的,只有碰到第一个非0才记录firstPos。

②对于题目给定的精度,如果输入的数字位数不够,要补0,在结尾处还要补\0使得字符串可靠结束。

③处理完毕后,对于firstPos和pointPos做比较,如果前者小,说明是大于1的数,pointPos - firstPos即为10的几次方;如果后者小,说明是小数,应该用firstPos - pointPos + 1,+1是因为firstPos越过了小数点,而小数点不能算作一位,注意的是这个次方是负值。

④比较相等时,如果基数部分一致、次方也一致,才算相等。

代码如下:

#include <stdio.h>
#include <string.h>
#define MAX 110
struct result{
	char d[MAX]; // 0.xxx部分
	int k; // 10的k次方
};

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) // 不能以0开头,否则忽略
			continue;
		else{
			if (firstPos == -1)
				firstPos = i; // 第一个非0数字的位置
			if (index < n)
			{
				if (index < strlen(a))
					r.d[index++] = a[i]; // 对于特定的精度,有数字则填入相应数字,没有则补0
				else
					r.d[index++] = '0';
			}
		}
	}
	r.d[index] = 0; // 在数字结尾加\0,防止越界
	if (pointPos == -1)
		pointPos = i; // 如果没有找到小数点,则小数点在最后,这是个纯整数
	if (pointPos - firstPos < 0) // 判断小数点与第一个非0数字的位置关系,计算10的几次方
		r.k = - (firstPos - pointPos - 1); // 负次方,例如0.015,pointPos = 1, firstPos = 3, 3 - 1 - 1 = 1, -1是因为多算了小数点进去,0.15*10^-1
	else
		r.k = pointPos - firstPos; // 正次方,例如21.25,pointPos = 2,firstPos = 0,2-0=2,0.2125*10^2
	if (index == 0){ // 如果index = 0,代表值为0,则每一位都写0,再加\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;
}



posted on 2015-07-27 16:34  张大大123  阅读(111)  评论(0编辑  收藏  举报

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