标签：syn while height 判断 math esc input only 题意
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 if there exist two pagodas standing erect, labelled j and k respectively, such that i=j+k or i=j?k
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
#include<iostream> #include<algorithm> #include<vector> #include<queue> #include<deque> #include<stack> #include<map> #include<set> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #define fuck(x) cout<<#x<<" = "<<x<<endl; using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 100086; const int inf = 2.1e9; const ll INF = 999999999999999; const double eps = 1e-6; int gcd(int a,int b) { return b?gcd(b,a%b):a; } int main() { // ios::sync_with_stdio(false); // freopen("in.txt","r",stdin); int T; scanf("%d",&T); int n,a,b; int cases = 0; while(T--){ cases++; scanf("%d%d%d",&n,&a,&b); int ans =n/gcd(a,b); fuck(ans) printf("Case #%d: ",cases); if(ans&1){printf("Yuwgna\n");} else printf("Iaka\n"); } return 0; }
标签：syn while height 判断 math esc input only 题意
原文地址：https://www.cnblogs.com/ZGQblogs/p/9743289.html
