Pagodas
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2045 Accepted Submission(s): 1405
Problem Description
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 aand 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. 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.
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.
For each test case, the first line provides the positive integer n (2≤n≤20000) and two different integers a and b.
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 12
Sample 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
Source
- 签到数论题
- 求gcd(a,b)=d，如果a和b互质则可以到达每个位置，否则总共到达位置数量为n/d
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <climits> 7 #include <cmath> 8 #include <vector> 9 #include <queue> 10 #include <stack> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL ; 15 typedef unsigned long long ULL ; 16 const int maxn = 1e5 + 10 ; 17 const int inf = 0x3f3f3f3f ; 18 const int npos = -1 ; 19 const int mod = 1e9 + 7 ; 20 const int mxx = 100 + 5 ; 21 const double eps = 1e-6 ; 22 const double PI = acos(-1.0) ; 23 24 int gcd(int x, int y){ 25 return y?gcd(y,x%y):x; 26 } 27 int T, n, a, b, c, d, Yuwgna; 28 int main(){ 29 // freopen("in.txt","r",stdin); 30 // freopen("out.txt","w",stdout); 31 while(~scanf("%d",&T)){ 32 for(int kase=1;kase<=T;kase++){ 33 scanf("%d %d %d",&n,&a,&b); 34 d=gcd(a,b); 35 c=(n/d)-2; 36 Yuwgna=c&1; 37 printf("Case #%d: %sn",kase,Yuwgna?"Yuwgna":"Iaka"); 38 } 39 } 40 return 0; 41 }
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