HDU5512 Pagodas (欧几里得)
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Pagodas
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2036 Accepted Submission(s): 1398
Problem Description
Two monks, Yuwgna and Iaka, decide to make glories great again. They take turns to build pagodas and Yuwgna takes first. For each turn, one can rebuild a new pagodas labelled
This is a game for them. The monk who can not rebuild a new pagoda will lose the game.
Input
The first line contains an integer t (1≤t≤500) which is the number of test cases.
For each test case, the first line provides the positive integern (2≤n≤20000) and two different integers a and b .
For each test case, the first line provides the positive integer
Output
For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time.
Sample Input
162 1 23 1 367 1 2100 1 28 6 89 6 810 6 811 6 812 6 813 6 814 6 815 6 816 6 81314 6 81994 1 131994 7 12
Sample Output
Case #1: IakaCase #2: YuwgnaCase #3: YuwgnaCase #4: IakaCase #5: IakaCase #6: IakaCase #7: YuwgnaCase #8: YuwgnaCase #9: IakaCase #10: IakaCase #11: YuwgnaCase #12: YuwgnaCase #13: IakaCase #14: YuwgnaCase #15: IakaCase #16: Iaka
Source
2015ACM/ICPC亚洲区沈阳站-重现赛(感谢东北大学)
思路:设c = gcd(a,b), 则设 a = k*c, b = p*c,则可以得出,不论是a-b或者b-a或者a+b,得到的结果都会是c的倍数,并且得到的最小的数为c,所以用n/c即可求出可以剪多少个灯塔。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t,a,b,n,cas=0;int gcd(int a, int b){ return b ? gcd(b,a%b) : a;}int main(){ scanf("%d",&t); while(t --){ scanf("%d%d%d",&n,&a,&b); int r = gcd(a,b); int x = n / r; printf("Case #%d: ",++cas); if(x % 2) printf("Yuwgna\n"); else printf("Iaka\n"); } return 0;}
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