Weird Clock Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1771 Accepted Submission(s): 643
Problem Description A weird clock marked from 0 to 59 has only a minute hand. It won't move until a special coin is thrown into its box. There are different kinds of coins as your options. However once you make your choice, you cannot use any other kind. There are infinite number of coins of each kind, each marked with a number d ( 1 <= d <= 1000 ), meaning that this coin will make the minute hand move d times clockwise the current time. For example, if the current time is 45, and d = 2. Then the minute hand will move clockwise 90 minutes and will be pointing to 15.
Now you are given the initial time s ( 1 <= s <= 59 ) and the coin's type d. Write a program to find the minimum number of d-coins needed to turn the minute hand back to 0.
Input There are several tests. Each test occupies a line containing two positive integers s and d.
The input is finished by a line containing 0 0.
Output For each test print in a single line the minimum number of coins needed. If it is impossible to turn the hand back to 0, output "Impossible".
Sample Input
30 1 0 0
Sample Output
1
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考慮無法達到題目要求，即輸出“Impossible” 的條件是當出現重複出先以前出現過的時間的時候即為無法達到要求，我設定了一個長度為60的數組記錄是否當前的時間是否已經出現過。
/**************************** *Name:Weird Clock.c *Tags:ACM water *Note:當第二次出現相同結果時表示不可能實現題目要求，輸出Impossible ****************************/#include <stdio.h>int main(){ int s, d, t, save[60], i; while(scanf("%d%d", &s, &d) != EOF && (s || d)) { for(i = 0; i < 60; i++) { save[i] = 0; } if(d == 0 && s) { printf("Impossible\n"); continue; } t = 0; while(s != 0 && !save[s]) { save[s] = 1; s = (s * (d + 1)) % 60; t++; } if(s == 0) { printf("%d\n", t); } else { printf("Impossible\n"); } } return 0;}
