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…

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)=…

http://acm.hdu.edu.cn/showproblem.php?pid=4861 两个人进行游戏,桌上有k个球,第i个球的值为1^i+2^i+⋯+(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. #include<iostream> #include<cstdio> using namespace std; int main() { int k,p,l; while(~scanf("%d%d",&k…

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<<"…

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…

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.....…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…