hide handkerchief

Time Limit: 10000/3000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)

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 對于題目的理解： 這個題目可以理解為下面的模型。 開始處于x的位置，每次隔M個訪問一次，問最后能否把原來的N個全部訪問。(不妨假設這N個序號是從0到N-1) 如果我們從開始一共走了K步，那么可以得到現在所處的位置是(x+k*M)%N。可以知道如果訪問到某個節點的時候，這個節點在前面已經出現過了，且這N個節點沒有全部訪問完的時候結果是不可能訪問完這N個節點。如果出現節點重復，此時恰好走了N步，那么就輸出yes。以上的想法可以簡單的畫圖理解。 由上述分析可知，如果(x+K*M)%N=x(表示出現重復節點)，且K=N時結果是yes，所以化簡這個式子有：(K*M)%N=0，那么要找到這個最小的K滿足這個式子自然會想到最小公倍數。再進一步想如果N*M是M和N的最小公倍數，那么M和N的最大公約數必然是1。也就有了下面的程序。 #include<iostream> #include<string> #include<string.h> using namespace std; int gcd(int a, int b) {if (a > b){a = a + b;b = a - b;a = a - b;}while (a != 0){int c = b%a;b = a;a = c;}return b; } int main() {int n, m;while (cin >> n >> m){if (n == -1 && m == -1)break;int temp= gcd(n, m);if (temp ==1)cout << "YES" << endl;else cout << "POOR Haha" << endl;}return 0; }

　　

轉載于:https://www.cnblogs.com/meow371/p/5175275.html

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