ACM刷题之HDU————hide handkerchief
2016-04-19 21:50
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hide handkerchief |
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
Total Submission(s): 6448 Accepted Submission(s): 2109 |
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 Input3 2 -1 -1 |
Sample OutputYES 算是之前刷HUD step的时候留下来的题目吧。当时没看懂。 直到我做完Wolf and Rabbit 的时候。发现这两个是一样的。 其实就是判断两个数是否是互质的数就好了。 下面是ac代码 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; int main() { __int64 a,b,t; while(scanf("%I64d%I64d",&a,&b)!=EOF) { if(a==b&&a==-1) break; if(a<b) { t=b; b=a; a=t; } while(a%b!=0) { t=a%b; a=b; b=t; } if(b==1) printf("YES\n"); else printf("POOR Haha\n"); } } |
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