hdu 4870 rating(高斯消元求期望)】的更多相关文章

Rating Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 872    Accepted Submission(s): 545Special Judge Problem Description A little girl loves programming competition very much. Recently, she h…
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4870 原题: Rating Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 654    Accepted Submission(s): 415 Special Judge Problem Description    A little…
Time travel Problem Description Agent K is one of the greatest agents in a secret organization called Men in Black. Once he needs to finish a mission by traveling through time with the Time machine. The Time machine can take agent K to some point (0…
Where is the canteen Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1070    Accepted Submission(s): 298 Problem Description After a long drastic struggle with himself, LL decide to go for some…
Time travel Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1480    Accepted Submission(s): 327 Problem Description Agent K is one of the greatest agents in a secret organization called Men in B…
Crazy Typewriter Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 391    Accepted Submission(s): 109 Problem Description There was a crazy typewriter before. When the writer is not very sober, it…
网络预选赛的题目……比赛的时候没有做上,确实是没啥思路,只知道肯定是整数分解,然后乘起来素数的幂肯定是偶数,然后就不知道该怎么办了… 最后题目要求输出方案数,首先根据题目应该能写出如下齐次方程(从别人那里盗的……): a11*x1  ^  a12*x2  ^  ...  ^  a1n*xn=0 a21*x1  ^  a22*x2  ^  ...  ^  a2n*xn=0 ... an1*x1  ^  an2*x2  ^  ...  ^  ann*xn=0,Aij表示选的第j个数的第i个质数(可能…
Agent K is one of the greatest agents in a secret organization called Men in Black. Once he needs to finish a mission by traveling through time with the Time machine. The Time machine can take agent K to some point (0 to n-1) on the timeline and when…
题目大意:给定一个数组,求这些数组通过异或能得到的数中的第k小是多少 首先高斯消元求出线性基,然后将k依照二进制拆分就可以 注意当高斯消元结束后若末尾有0则第1小是0 特判一下然后k-- 然后HDU输出long long是用%I64d 不管C艹还是G艹都是 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define M 10100 using namesp…
3949冰上走 题意: 给你 N个数,从中取出若干个进行异或运算 , 求最后所有可以得到的异或结果中的第k小值 N个数高斯消元求出线性基后,设秩为$r$,那么总共可以组成$2^r$中数字(本题不能不选,所以$2^r -1$) 然后如果$k \ge 2^r$就不存在啦 否则一定可以有$k$小,因为现在$1..r$行每行都有一位是1(左面是最高位) 从高到低枚举k的二进制,如果是1就异或上对应的行就行了,最后就是k小值啦 #include <iostream> #include <cstdi…