题目链接:hdu 4861 Couple doubi 题目大意:两个人进行游戏,桌上有k个球,第i个球的值为1i+2i+⋯+(p−1)i%p,两个人轮流取,假设DouBiNan的值大的话就输出YES,否则输出NO. 解题思路: 首先是DouBiNan先取,所以肯定优先选取剩余中值最大的,于是不存在说DouBiNan值小的情况,仅仅有大于和小于. 然后,对于val(i)=1i+2i+⋯+(p−1)i%p来说,仅仅有当i=ϕ(p)=p−1(p为素数)时,val(i)=p−1,其它情况下val(i)=…

Couple doubi Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4861 Description DouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of th…

Couple doubi 题目链接: http://acm.hust.edu.cn/vjudge/contest/121334#problem/D 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 th…

题目链接 题意 : 有K个球,给你一个数P,可以求出K个值,(i=1,2,...,k) : 1^i+2^i+...+(p-1)^i (mod p).然后女朋友先取,再xp取,都希望赢,如果女朋友能赢输出YES,否则输出NO 思路 :这个题,在纸上算算差不多就出来结果了,因为要赢,所以一开始必定拿大的,根据规律可以发现最后的那个取余结果不是0就是某个数,所以就看那个数有奇数个还是偶数个即可. 官方题解: #include <stdio.h> #include <string.h> #…

题目链接 可以瞎搞一下,找找规律 题意:两个人进行游戏,桌上有k个球,第i个球的值为1i+2i+⋯+(p−1)i%p,两个人轮流取,如果DouBiNan的值大的话就输出YES,否则输出NO. 分析:解题报告 #include <cstdio> #include <iostream> using namespace std; int main() { int k, p; while(cin>>k>>p) { )&) cout<<"…

HDU 1005 Number Sequence(数论) Problem Description: A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multipl…

Least Common Multiple (HDU - 1019) [简单数论][LCM][欧几里得辗转相除法] 标签: 入门讲座题解 数论 题目描述 The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7…

七夕节 (HDU - 1215) [简单数论][找因数] 标签: 入门讲座题解 数论 题目描述 七夕节那天,月老来到数字王国,他在城门上贴了一张告示,并且和数字王国的人们说:"你们想知道你们的另一半是谁吗?那就按照告示上的方法去找吧!" 人们纷纷来到告示前,都想知道谁才是自己的另一半.告示如下: 数字N的因子就是所有比N小又能被N整除的所有正整数,如12的因子有1,2,3,4,6. 你想知道你的另一半吗? Input 输入数据的第一行是一个数字T(1<=T<=500000)…

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)…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4861 解题报告:桌子上有k个球 ,第i个球的价值wi = 1^i+2^i+...+(p-1)^i (mod p),现在两个人轮流取,如果第一个取的人最后得到的价值总和最大,输出YES,否则 输出NO . k和p的范围都很大,如果都算出来那是不可能的,所以一定是有规律的,所以我把前面的几个打表看了一下,果然有规律,第1.2.3.......(p-2),(p-1)个球的价值分别是: 0,0,0.....…