题意:给出n对点a,b 要求从没对点中选出一个,且最终选出的点n个数不能存在相同的.输入数据满足每种数最多出现3次,最少出现1次 思路:第i对点的编号2*i, 2*i+1, 因为每个数最多出现3次,那么完全可以枚举每个数,然后相同的数之间的编号建立关系(¬a Λ ¬b 为真,表示这两个编号不能同时选), 然后同一对的俩编号之间也有关系(a xor b为真,代表a和b必须选且只能选一个,a xor b 可以写成 (a V b) Λ (¬a V ¬b)),这样跑完twosat就能得到一个满足情…
Description Misha trains several ACM teams at the university. He is an experienced coach, and he does not underestimate the meaning of friendly and collaborative atmosphere during training sessions. It used to be that way, but one of the teams happen…
不要62 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 26743 Accepted Submission(s): 9385 Problem Description 杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer).杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可以…
题意:有n杯盐溶液,给定每杯里面盐的质量以及盐溶液的质量.问你有多少种方案选择一个子集,使得集合里面的盐溶液倒到一个被子里面以后,浓度为A/B. 折半枚举,暴力搜索分界线一侧的答案数,跨越分界线的答案,一侧用map存下来,枚举另一侧去统计. #include<cstdio> #include<map> using namespace std; typedef long long ll; ll ans; int n,a,b,m[39],t[39]; map<ll,int>…
题解及证明:http://www.cnblogs.com/StupidBoy/p/5241258.html #include<cstdio> #include<algorithm> using namespace std; int n,m,a[100100],b[100100]; int main() { //freopen("h.in","r",stdin); scanf("%d%d",&n,&m); f…
Few people know, but a long time ago a developed state existed on Mars. It consisted of n cities, numbered by integers from 1 to n, the capital had the number 1. Some pairs of cities were connected by a road. The residents of the state were very prud…
暴力搞肯定不行,因此我们从小到大枚举素数,用n去试除,每次除尽,如果已经超过20,肯定是no.如果当前枚举到的素数的(20-已经找到的质因子个数)次方>剩下的n,肯定也是no.再加一个关键的优化,如果剩下的次数是1了,就直接判定剩下的n是否是素数.这样可以保证次方>=2,将我们需要枚举的素数限制在200w以内,就可做了.线性筛在这题虽然不必要,但是可以当个板子存下来. The hacker Michael develops breakthrough password manager, whic…
