标签：
| Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 5029 Accepted Submission(s): 1670 |
|
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". |
|
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 |
|
Source
HDU 2007-6 Programming Contest
|
|
Recommend
xhd
|
思路：此题虽然是水题，但是还是学到不少东西，尤其是最大公约数的求法与证法，刚开始想开辟一个非常大的数组的但是占用内存太大，应该理解好题意再进行求解。
最大公约数为1的情况。
1 #include<stdio.h> 2 3 int GCD(int a ,int b) 4 { 5 return b==0?a:GCD(b,a%b); 6 } 7 //查找是否有最大公约数 8 int main(int argc, char *argv[]) 9 { 10 int N,M; 11 while(scanf("%d%d",&N,&M)) 12 { 13 if(N==-1&&M==-1) 14 break; 15 int temp=GCD(N,M); 16 if(temp==1) 17 printf("YES\n"); 18 else 19 printf("POOR Haha\n"); 20 } 21 return 0; 22 }
标签：
原文地址：http://www.cnblogs.com/GoFly/p/4239872.html
