原題地址： http://acm.hdu.edu.cn/game/entry/problem/show.php?chapterid=1§ionid=2&problemid=6
hide handkerchief |
| Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 1708 Accepted Submission(s): 609 |
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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 . Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A. So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me "YES", else tell me "POOR Haha". |
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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.
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Output
For each input case, you should only the result that Haha can find the handkerchief or not.
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Sample Input
3 2 -1 -1 |
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Sample Output
YES |
首先這題開始沒看出來什麼道道，就打算用遍歷枚舉的方法做，結果是堆棧溢出，數組開得太大了，於是乎不知道怎麼辦，所以我百度了一下，的值這題的規律是M與N互質，所以就寫了這個代碼~~貼上
#include<iostream>
using namespace std;
int main()
{
int m, n, temp;
while (cin>>m>>n)
{
if ((m == -1)&&(n == -1))
break;
while (n != 0)
{
temp = n;
n = m % n;
m = temp;
}
if (m == 1) cout<<"YES"<<endl;
else cout<<"POOR Haha"<<endl;
}
system ("pause");
return 0;
}
其實這題本身沒有什麼講的，關鍵是我又踏回了編程之路……還有決定這段時間學習數論，感覺挺有意思的。