1060 Are They Equal (25point(s)) - PAT 甲級 C語言

1060 Are They Equal (25point(s))

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 10​100​​, 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.d[1]…d[N]*10^k (d[1]>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

題目大意：

輸入一個整數 N，兩個小數 A、B，問將這兩個小數保存爲 N 位小數的科學計數法後，兩小數是否相等，並輸出用科學計數法表示後的數

設計思路：

重點在於確定係數的有效數字和指數

• 去除小數點前的前導 0 後
  • 若小數點前有數字，指數爲正
    • 則數字有幾個，對應指數就爲幾
    • 保留有效位數
  • 若小數點前無數字，指數爲負
    • 則小數點後有幾個 0，對應指數就爲負幾
    • 保留有效位數
• 有效位數不足題目要求，後面補 0

這裏提供一些測試用例，

1 0 00.0
1 0.1 1.0
2 0.0 0.0010
4 00012.3 0.00123

注意：

• 指數爲 0 和 1 次方也要輸出
• 無需四捨五入

編譯器：C (gcc)

#include <stdio.h>

int main(void) {
        int n;
        char num[2][110] = {0};
        char ch;
        int i, flag, flag2, count, exp[2] = {0};

        scanf("%d ", &n);
        for (i = 0; i < 2; i++) {
                flag = 0;
                flag2 = 0;
                count = 0;
                exp[i] = 0;
                while ((ch = getchar()) != ' ' && ch != '\n') {
                        if (flag == 0) {
                                if (ch == '.') {
                                        flag = -1;
                                        continue;
                                } else if (ch > '0') {
                                        flag = 1;
                                }
                        }
                        if (flag == -1) {
                                if (ch != '0') {
                                        flag2 = 1;
                                }
                                if (flag2 == 0) {
                                        exp[i]--;
                                } else if (flag2 == 1 && count < n) {
                                        num[i][count] = ch;
                                        count++;
                                }
                        }
                        if (flag == 1) {
                                if (ch != '.') {
                                        if (flag2 == 0) {
                                                exp[i]++;
                                        }
                                        if (count < n) {
                                                num[i][count] = ch;
                                                count++;
                                        }
                                } else {
                                        flag2 = 1;
                                }
                        }
                }
                if (flag == -1 && flag2 == 0) {
                        exp[i] = 0;
                }
                while (count < n) {
                        num[i][count] = '0';
                        count++;
                }
        }

        if (exp[0] == exp[1] && strcmp(num[0], num[1]) == 0) {
                printf("YES 0.%s*10^%d", num[0], exp[0]);
        } else {
                printf("NO 0.%s*10^%d 0.%s*10^%d", num[0], exp[0], num[1], exp[1]);
        }

        return 0;
}