hoj 1006

Weird Clock

My Tags	(Edit)

	Source : ZJU
	Time limit : 3 sec		Memory limit : 32 M

Submitted : 9417, Accepted : 1821

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 ( 0 <= 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 ( 0 <= 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

有必要说下的是这种标记方法 被访问的用 1 标记 ， 没访问的用 0 标记； 数数组去记录标记的位置

算法思想：

         第一步： 数据结构：要想到使用一个数组来标记 鈡所走的位置 1表示走过了 0 表示还没有走过

         第二步： ：算法求其余数 在一个while循环里面完成 知道余数为0 停止 见代码 while循环中 这里最关键

         第三遍 ：打印就可以了

源代码：

    WA 的不知道为什么 WA  

#include
#include
using namespace std;

int main()
{
   int a, b;
   int c[61];
   while(cin>>a>>b)
   {
       memset(c, 0, sizeof(c));
       if(a == 0 && b == 0)
       {
           return 0;
       }
       if(a == 0)
       {
           cout<<"0"<

再来一个 AC 的源代码：

#include 

int main()
{
    int s, p;
    
    while(scanf("%d %d", &s, &p) == 2){
        if((s == 0) && (p == 0)){
            return 0;
        }else if((s == 0) && (p != 0)){
            printf("0\n");
        }else if(s*(p+1) % 60 == 0){
            printf("1\n");
        }else if(s*(p+1)*(p+1) % 60 == 0){
            printf("2\n");
        }else{
            printf("Impossible\n");
        }
    }
    
    return 0;
}

这里要说的是 最多使用两个金币 2 到二就可以停止了 ，为什么了 。。。。。好吧不知道。。。。