Pagodas(等差数列)
2015-11-01 19:52
405 查看
PagodasTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 24 Accepted Submission(s): 22 [align=left]Problem Description[/align] 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. [align=left]Input[/align] 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. [align=left]Output[/align] For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time. [align=left]Sample Input[/align] 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 [align=left]Sample Output[/align] 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 |
现在给出一个博弈局面,当某个人不能再建造塔时为输。问你谁能赢。
思路:当且仅当 a和b处于某个等差数列(差值不为1)时,才无法使得所有位置都建上塔。相反,则n个位置均可建塔。等差数列的差值 d = gcd(a, b),求出可以建塔的个数就可以了。
代码:
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int gcd(int x,int y){ return !y?x:gcd(y,x%y); } int main(){ int T,flot=0; int n,a,b; scanf("%d",&T); while(T--){ scanf("%d%d%d",&n,&a,&b); int t=gcd(a,b); //printf("%d\n",t); int temp=n/t-2; if(temp&1)printf("Case #%d: Yuwgna\n",++flot); else printf("Case #%d: Iaka\n",++flot); } return 0; }
相关文章推荐
- Django form 文件上传
- Leetcode Algorithm No.268 Missing Number
- ubuntu,django-admin.py
- HDU_4780_Candy Factory(最小费用流)
- 使用wagon-maven-plugin插件自动部署项目
- ACM ICPC Vietnam National Second Round I. Space Tour(预处理 + 枚举)
- 美团云服务器django web平台搭建全过程
- hdu 5512 Pagodas
- 【学习笔记0001】Go初学者需要关注的50个注意事项
- 跟着猫哥学Golang[5] - 循环
- django登陆验证装饰器
- go语言实现设计模式:前言
- go语言实现设计模式(二):简易工厂
- HDU 5512 Pagodas (2015沈阳现场赛,找规律+gcd)
- HDU-5512 Pagodas(GCD)
- 创作与google新logo相似的个人logo
- [Django与表单]写一个简单的Django form表单处理
- 第四章:Django 的模板系统(转)
- django 简易博客开发 1 安装、创建、配置、admin使用(转)
- hdoj4722Good Numbers【找规律】