HDU 5512(博弈论)
n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from
1 to
n. However, only two of them (labelled
a and
b, where
1≤a≠b≤n) withstood the test of time.
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 i (i∉{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled j and k respectively, such that i=j+k or i=j−k
For each test case, the first line provides the positive integer n (2≤n≤20000) and two different integers a and b. Output For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time. Sample Input
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 i (i∉{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled j and k respectively, such that i=j+k or i=j−k
. Each pagoda can not be rebuilt twice.
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.
This is a game for them. The monk who can not rebuild a new pagoda will lose the game.
For each test case, the first line provides the positive integer n (2≤n≤20000) and two different integers a and b. Output For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time. Sample Input
16 2 1 2 3 1 3 67 1 2 100 1 2 8 6 8 9 6 8 10 6 8 11 6 8 12 6 8 13 6 8 14 6 8 15 6 8 16 6 8 1314 6 8 1994 1 13 1994 7 12Sample Output
Case #1: Iaka Case #2: Yuwgna Case #3: Yuwgna Case #4: Iaka Case #5: Iaka Case #6: Iaka Case #7: Yuwgna Case #8: Yuwgna Case #9: Iaka Case #10: Iaka Case #11: Yuwgna Case #12: Yuwgna Case #13: Iaka Case #14: Yuwgna Case #15: Iaka Case #16: Iaka
转载来自：https://blog.csdn.net/mystery_guest/article/details/52072810
emmm昨天做这题的时候没有想到gcd被题目的样例给误导了 一直想的是他们的差值的多少却没有想到其他的
今天看了下题解是扩展欧几里得
意思就是如果ab互为质数的话那么一定可以得到一个1 那么一定也可以得到每个东西或者a==1||b==1
如果他们不互相为质数的话 那么他们可以取到他们的最大公约数里面所有有关公约数的倍数
就是这个样子那么就是n的个数除以他们的最大公约数就是里面可能的个数 之后减不减2都无所谓
代码：
#include
using namespace std;
int main()
{
int t,n,a,b;cin>>t;
for(int i=1;i<=t;i++)
{
scanf("%d%d%d",&n,&a,&b);
int temp = __gcd(a,b);
printf("Case #%d: ",i);
if(a==1||b==1||temp==1)//如果ab互质
{
if(n%2==1)printf("Yuwgna\n");
else printf("Iaka\n");
}else //如果ab不互质 那么只要是gcd都能修建
{
int gg=n/temp-2;
if(gg%2==1)printf("Yuwgna\n");
else printf("Iaka\n");
}
}
return 0;
}