解题报告
两人轮流取球，大的人赢，，，
贴官方题解，，，反正我看不懂。，，先留着理解
找规律找到的，，，sad，，，
非常easy找到循环节为p-1，每个循环节中有一个非零的球。所以仅仅要推断有多少完整循环节。在推断奇偶，。，
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int main()
{
int k,p,i,j;
while(scanf("%d%d",&k,&p)!=EOF){
int q=k/(p-1);
if(q%2==0)
printf("NO\n");
else printf("YES\n");
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath> using namespace std; int main()
{
int k,p,i,j;
while(scanf("%d%d",&k,&p)!=EOF){
int q=k/(p-1);
if(q%2==0)
printf("NO\n");
else printf("YES\n");
}
return 0;
}
Couple doubi
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 145 Accepted Submission(s): 122
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”.
(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).
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
Source