hdoj 4861 Couple doubi 【打表找规律】_doubixp doubinan-优快云博客

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/50640590

本文介绍了一个涉及数学博弈的问题，通过分析球的价值计算方法和游戏规则，探讨了在特定条件下先手玩家是否能确保胜利的策略。该问题与数论中的费马小定理相关，并提供了一种简洁的解决方案。

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Couple doubi

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1467 Accepted Submission(s): 1018

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

题意：有k个球，第i个球的价值 = 1^i + ... + (p-1)^i mod p。有两个人轮流取球，最后比较价值和，价值和大的赢。问先手有没有希望赢。

思路：打表。。。不过应该是一道和费马小定理有关的数论推导。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (100000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
//LL Pow(LL a, LL n, LL p)
//{
//    LL ans = 1LL;
//    while(n)
//    {
//        if(n & 1)
//            ans = ans * a % p;
//        a = a * a % p;
//        n >>= 1;
//    }
//    return ans;
//}
//int main()
//{
//    // p = 3  0 2 0 2 0 2 ......
//    // p = 5  0 0 0 4 0 0 0 4 ......
//    // p = 7  0 0 0 0 0 6 ......
//    int k = 10;
//    for(int i = 1; i <= 10; i++)
//    {
//        LL ans = 0;
//        for(int j = 1; j <= 6; j++)
//            ans = (ans + Pow(j, i, 7)) % 7;
//        printf("%lld ", ans);
//    }
//    return 0;
//}
int main()
{
    int k, p;
    while(scanf("%d%d", &k, &p) != EOF)
    {
        bool flag = (k / (p-1)) & 1;
        printf(flag ? "YES\n" : "NO\n");
    }
    return 0;
}