C. The Game Of Parity time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n cities in Westeros. The i-th city is inhabited by ai people. Daenerys and Stannis play the following game:…

如果sm[j]和sm[i]奇偶性相同,那么(i+1,j)个数为偶数如果奇偶性相同看成是朋友,不同的看成是敌人,那么就跟bzoj1370的做法差不多了. 如果奇偶性相同,就将x和y合并,x+n,y+n合并 如果奇偶性不同,就将x和y+n合并,y和x+n合并 #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> using namespace std; #define re…

C. Parity Game 题目连接: http://www.codeforces.com/contest/298/problem/C Description You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes…

Parity game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6816   Accepted: 2636 Description Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You choose a conti…

1003 看篇国家论文 <从<parity>的解法谈程序优化> 对于区间i,j 如果用sum[i],sum[j]来表示到i的1的个数的奇偶性 那么仔细想下 sum[i-1] 若与区间i,j相等 则sum[j]为偶 否则为奇 那么就可以把性质相同的合并在一个集合里 性质相同为朋友 不同为敌人 可以把一个端点分成两个 一个是自己一个是他的敌人 当与别的点合并时根据朋友的朋友是朋友 朋友的敌人是敌人 敌人的敌人 是朋友 这些原则 来进行合并 ,并判断是不是有矛盾 端点比较大 用map离散…

Parity game 题意:一个长度为N(N < 1e9)内的01串,之后有K(K <= 5000)组叙述,表示区间[l,r]之间1的个数为odd还是even:问在第一个叙述矛盾前说了几句话? Sample Input 10 N 5 K 1 2 even 3 4 odd 5 6 even 1 6 even (错误) 7 10 odd Sample Output 3 思路:对于判断正误的问题,一定要知道什么情况会导致错误.如样例:当前输入的区间[1,6]的端点均包含于前面输入的端点中时(包含并…

Even Parity We have a grid of size N x N. Each cell of the grid initially contains a zero(0) or a one(1). The parity of a cell is the number of 1s surrounding that cell. A cell is surrounded by at most 4 cells (top, bottom, left, right). Suppose we h…

http://acm.hdu.edu.cn/showproblem.php?pid=2700 题目意思很重要:  //e:是要使字符串中1的个数变成偶数.o:是要使字符串中1的个数变成奇数 Parity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1855    Accepted Submission(s): 1447 Proble…

The Game Of Parity Solution: 这个题只需要分类讨论就可以解决. 先分别统计奇数和偶数的个数. 然后判断谁走最后一步,如果走最后一步时候同时有偶数和奇数,那么走最后一步的赢.如果没有呢,就看剩下的是奇数还是偶数,是偶数不用说,是奇数看剩奇数个还是偶数个. 针对这几种情况讨论一下就可以. #include <bits/stdc++.h> using namespace std; int n, k, o, e; int main() { cin >> n &g…

Problem Description A bit string has odd parity if the number of 1's is odd. A bit string has even parity if the number of 1's is even.Zero is considered to be an even number, so a bit string with no 1's has even parity. Note that the number of 0's d…