杭電ACM-2104 hide handkerchief
hide handkerchief
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6917 Accepted Submission(s): 3110
Problem Description
The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.Now I introduce the game to you. Suppose there are N people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes .
Input
There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data.
Output
For each input case, you should only the result that Haha can find the handkerchief or not.
Sample Input
3 2 -1 -1
Sample Output YES
開始看了半天沒看懂題目，發現是N個人圍成一個圈，定義M，每隔M個人搜尋一次，類似於約瑟夫環但又不去掉盒子的搜尋，當M與N互質是必然搜到手帕。用輾轉相除法求得二者的公約數為1是成立。
AC程式碼如下：
#include <iostream>
using namespace std;
int main()
{
int a ,b ,r ;
while (cin >> a >> b )
{
if(a==-1&&b==-1) break;
else
while(b!=0 )
{
r = a%b;
a = b;
b = r;
}
if(a==1)
{
cout<<"YES"<<endl;
}
else
cout<<"POOR Haha"<<endl;
}
return 0 ;
}