PAT甲级1060. Are They Equal

野性酷女 2022-06-11 07:11 126阅读 0赞

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

### 题目要求 ###

将两个不超过10^100的浮点数按照输入输入的精确度表示，然后比较两个浮点数是否相同，如果相同则输出 YES，并将使用科学计数法表示的数字输出；如果不相同，则输出NO，并将两个数字以科学计数法输出；

### 解题思路 ###

题目要求不超过10^100，那么这个浮点数只能用数组来表示；数字大小的比较可以通过比较其所对应的的ASCII，即不用char来表示每一个数字，通过字符的比较即可得到数字是否相同;

void getRidZero() 这一个函数用于将num\[\]数组清除前导零和小数点，然后将剩下的复制到数组pNum()中；同时将小数点的位置以及给予的浮点数第一个非零的数字记录在decimalPoint和index中；

void formatPrint()这一个函数用于将浮点数格式化输出；其中decimalPoint-index表示的是科学计数法的指数；decimalPoint1-index>=0说明原来的数大于0，如：123.2223；如果decimalPoint1-index<=0则说明是纯小数，如：0.00123，它的index=4,decimalPoint=1,小数点占了一位，因此指数需要将decimalPoint-index的值+1；

对于两个0比较，如：0000和000.000则先判断是否都为0，然后再输出；

数组的大小开到10000，是开始时开到100，发现没有AC，是因为题目给出的是数字位数不超过100，但数字的前导零的个数是不确定的；

需要注意的要点：结尾需要使用0进行补足

### C语言代码 ###

#include <stdlib.h>
    #include <stdio.h>
    #include <stdbool.h>
    #include <string.h>
    
    //用于将数组的前导零和小数点去掉，并将num[]数组的其他字符保存在 pNUm[]中
    void getRidZero(char  num[10000], int *index, int *decimalPoint, char  pNum[10000])
    {
        //做一个循环除去开头的零和小数点
        for (int i = 0; i < 10000; i++)
        {
            if (num[i] != '0'&&num[i] != '.'&&num[i] != '\0')
            {
                *index = i;
                break;
            }
        }
        //用于获取小数点的位置
        for (int i = 0; i < 10000; i++)
        {
            if (num[i] == '\0' || num[i] == '.')
            {
                *decimalPoint = i;
                break;
            }
        }
    
        int j = 0;
    
        for (int i = *index; i < 10000; i++)
        {
            if (num[i] == '\0' || num[i] < 0)
            {
                pNum[j] = '0';
                j++;
            }
            else if (num[i] == '.')
            {
    
                continue;
            }
    
            else
            {
                pNum[j] = num[i];
                j++;
            }
    
        }
    
    }
    
    //用于将pNum[]中保存的数按照科学计数法输出
    void formatPrint(int accuracy, char  pNum[10000], int decimalPoint, int index)
    {
        printf("0.");
        for (int i = 0; i < accuracy; i++)
        {
            printf("%C", pNum[i]);
        }
        if (decimalPoint - index >= 0)
        {
            printf("*10^%d", decimalPoint - index);
        }
        else
        {
            printf("*10^%d", decimalPoint - index + 1);
        }
    }
    
    int main() {
    
        int  accuracy;
        char num1[10000];
        char num2[10000];
    
        char pNum1[10000];
        char pNum2[10000];
    
        //用于记录小数点的位置
        int decimalPoint1 = 0;
        int decimalPoint2 = 0;
    
        //用于记录数字开始的位置
        int index1 = 0;
        int index2 = 0;
    
        bool equal = true;
    
        scanf("%d %s %s", &accuracy, num1, num2);
        getRidZero(num1, &index1, &decimalPoint1, pNum1);
        getRidZero(num2, &index2, &decimalPoint2, pNum2);
    
    
        for (int i = 0; i < accuracy; i++)
        {
            if (pNum1[i] != pNum2[i])
            {
                equal = false;
            }
        }
    
        if ((index1 - decimalPoint1) != (index2 - decimalPoint2))
        {
            equal = false;
        }
    
        bool notZero = false;
        //判断00000000 和000000.000000000
        for (int i = 0; i < 5000; i++)
        {
            if (pNum1[i] != '0' || pNum2[i] != '0')
            {
                //证明不是0
                notZero = true;
                break;
            }
        }
    
    
        if (!notZero)  //证明都是0 然后输出 0.0000*10^0 ，是第六个测试点
        {
            printf("YES 0.");
            for (int i = 0; i < accuracy; i++)
            {
                printf("%C", '0');
            }
            printf("*10^%d", 0);
    
        }
        else if (equal)    //如果两个数相等
        {
            printf("YES ");
    
            formatPrint(accuracy, pNum1, decimalPoint1, index1);
    
        }
        else          //如果两个数不想等
        {
            printf("NO ");
            formatPrint(accuracy, pNum1, decimalPoint1, index1);
    
            //输出第二个数
            printf(" ");
            formatPrint(accuracy, pNum2, decimalPoint2, index2);
        }
        system("pause");
        return 0;
    }

### 测试用例 ###

*  5　0000.0000123　0.0000123
 *  4　000123　　　　　00123.005
 *  4　00000　　　　　000.000001
 *  4　00.000　　　　　00000
 *  4　0.01　　　　　　0.01000005