Couple doubi(费马小定理或打表找规律+简单博弈)
2015-09-28 13:29
423 查看
Link:http://acm.hdu.edu.cn/showproblem.php?pid=4861
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1434 Accepted Submission(s): 993
Problem Description
DouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of this game is as following. There are k balls on the desk. Every ball has a value and the value of ith (i=1,2,...,k) ball is 1^i+2^i+...+(p-1)^i
(mod p). Number p is a prime number that is chosen by DouBiXp and his girlfriend. And then they take balls in turn and DouBiNan first. After all the balls are token, they compare the sum of values with the other ,and the person who get larger sum will win
the game. You should print “YES” if DouBiNan will win the game. Otherwise you should print “NO”.
Input
Multiply Test Cases.
In the first line there are two Integers k and p(1<k,p<2^31).
Output
For each line, output an integer, as described above.
Sample Input
Sample Output
Author
FZU
Source
2014 Multi-University Training Contest 1
Couple doubi
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1434 Accepted Submission(s): 993
Problem Description
DouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of this game is as following. There are k balls on the desk. Every ball has a value and the value of ith (i=1,2,...,k) ball is 1^i+2^i+...+(p-1)^i
(mod p). Number p is a prime number that is chosen by DouBiXp and his girlfriend. And then they take balls in turn and DouBiNan first. After all the balls are token, they compare the sum of values with the other ,and the person who get larger sum will win
the game. You should print “YES” if DouBiNan will win the game. Otherwise you should print “NO”.
Input
Multiply Test Cases.
In the first line there are two Integers k and p(1<k,p<2^31).
Output
For each line, output an integer, as described above.
Sample Input
2 3 20 3
Sample Output
YES NO
Author
FZU
Source
2014 Multi-University Training Contest 1
/*附上打表找规律代码: #include<stdio.h> #include<math.h> int main () { int i ,j ,sum; printf("9 3\n"); for(i = 1 ;i <= 9 ;i ++) { sum = 0; for(j = 1 ;j < 3 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 3; printf("%d " ,sum % 3); } printf("\n2 3\n"); for(i = 1 ;i <= 2 ;i ++) { sum = 0; for(j = 1 ;j < 3 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 3; printf("%d " ,sum % 3); } printf("\n9 2\n"); for(i = 1 ;i <= 9 ;i ++) { sum = 0; for(j = 1 ;j < 2 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 2; printf("%d " ,sum % 2); } printf("\n10 3\n"); for(i = 1 ;i <= 10 ;i ++) { sum = 0; for(j = 1 ;j < 3 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 3; printf("%d " ,sum % 3); } printf("\n10 5\n"); for(i = 1 ;i <= 10 ;i ++) { sum = 0; for(j = 1 ;j < 5 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 5; printf("%d " ,sum % 5); } printf("\n10 7\n"); for(i = 1 ;i <= 10 ;i ++) { sum = 0; for(j = 1 ;j < 7 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 7; printf("%d " ,sum % 7); } printf("\n10 11\n"); for(i = 1 ;i <= 10 ;i ++) { sum = 0; for(j = 1 ;j < 11 ;j ++) sum = sum + int (pow(j * 1.0 ,i * 1.0)) % 11; printf("%d " ,sum % 11); } getchar(); }*/
/*编程思想:多试几个很容易可以发现规律。。。只有在p-1的倍数时有非0元素存在。 那么由于是博弈,大家每次肯定都是选那个比较大的数,所以,若是 k/(p-1)为奇数则先手必胜,否则平局*/ //AC CODE: #include <iostream> #include <cstdio> using namespace std; int main() { int k, p; while(~scanf("%d%d", &k, &p)) { if((k/(p-1))%2==1) printf("YES\n"); else printf("NO\n"); } return 0; }
相关文章推荐
- 《C算法》笔记11:BST再平衡
- 多彩的Console打印新玩法
- linux下安装redis
- iscsi详解&&配置
- 【BLE】CC2541之动态广播
- 94 数据结构和方向感言
- Java习题
- USB驱动——键盘,U盘
- 临汾的前世今生
- 好团队不可能凭空出现,赢在Leader的可行规划
- win7系统eclipse下切换SVN用户
- SQL 设置自增,和default
- Java基础知识强化之IO流笔记01:异常的概述和分类
- Lab 4: Cache Geometries
- 灰盒测试
- 交叉编译Busybox,使开发板串口终端ls命令支持显示中文
- tableview--Grouped 模式 去掉 顶部空白
- JavaScript通告/订阅的例子
- EasyARM-iMX257如何配置出低速率CAN
- [leetcode] 287.Find the Duplicate Number