hdu 5512 Pagodas gcd()

Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1137    Accepted Submission(s): 808

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.

 

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.

 

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）

 

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以前好像做过，不过当时不是队友做的，就是瞎搞的，反正应该没弄清楚。

现在的思路是：

首先要确定区间内的数有多少个可以选，判断是奇数还是偶数，奇数Yuwgna胜，偶数Iaka胜。

给我的感觉是不断相减(辗转相除)找到最小的量，

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;int n;int cal(int a,int b){    if(b==0)  return a;    return cal(b,a%b);}int main(){   std::ios::sync_with_stdio(false);   int a,b,T,kase=0;cin>>T;   while(T--)   {       cin>>n>>a>>b;       if(a>b) swap(a,b);       int p=cal(b,a);       int num=   (a-1)/p+ (b-a)/p+1+ (n-b)/p;       printf("Case #%d: ",++kase);       puts(num&1?"Yuwgna":"Iaka");   }   return 0;}

代码修改后：

看了别人的代码后，直接n/p即可，因为p就是求gcd(a,b)。

两个操作 a+b和b-a，都有gcd|a+b,gcd|b-a,所以不管加减操作多少次都是gcd的倍数。

所以就是要求[1,n]内有多少个数是gcd的倍数。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;int n;int cal(int a,int b){    return b==0?a:cal(b,a%b);}int main(){   std::ios::sync_with_stdio(false);   int a,b,T,kase=0;cin>>T;   while(T--)   {       cin>>n>>a>>b;       int p=cal(a,b);       int num=   n/p;       printf("Case #%d: ",++kase);       puts(num&1?"Yuwgna":"Iaka");   }   return 0;}