一道ACM问题（数论）Problem DescriptionDouBiXp 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 valu
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一道ACM问题（数论）Problem DescriptionDouBiXp 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 valu
一道ACM问题（数论）
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
一道ACM问题（数论）Problem DescriptionDouBiXp 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 valu
#include <iostream>
using namespace std;
int main() {
int k, p;
while (cin >> k >> p) {
if (k / (p - 1) % 2) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
}
思路
当 i 是 p-1 的倍数的时候, 第i个小球的价值是p-1否则小球的价值是0