Mex Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 623    Accepted Submission(s): 209 Problem Description Mex is a function on a set of integers, which is universally used for impartial game t…

就是不知道时间该怎么处理,想了好久,看了别人的题解发现原来是暴力,暴力也很巧妙啊,想不出来的那种  -_-! #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; #define MOD 10000…

Flyer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 386    Accepted Submission(s): 127 Problem Description The new semester begins! Different kinds of student societies are all trying to adver…

Save Labman No.004 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 624    Accepted Submission(s): 154 Problem Description Due to the preeminent research conducted by Dr. Kyouma, human beings hav…

Zhuge Liang's Mines Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 239    Accepted Submission(s): 110 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous an…

Caocao's Bridges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 194    Accepted Submission(s): 89 Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But…

Save Labman No.004 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 979    Accepted Submission(s): 306 Problem Description Due to the preeminent research conducted by Dr. Kyouma, human beings ha…

The Donkey of Gui Zhou Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 389    Accepted Submission(s): 153 Problem Description There was no donkey in the province of Gui Zhou, China. A trouble m…

突然想到的节约时间的方法,感觉6翻了  给你n个数字,接着m个询问.每次问你一段区间内不大于某个数字(不一定是给你的数字)的个数 直接线段树没法做,因为每次给你的数字不一样,父节点无法统计.但是离线一下,如果后面的询问可以用前面已经处理过的一些东西,则可以节约时间.换句话说,就是直接把给数字z进行从小到大排序,每次暴力更新数字a小于等于数字z的叶子节点,数字a赋值为极大值,这样前面更新了的数字a在后面就不需要再更新了,而且后面的数字z一定不小于前面更新了的数字a,而且总数一共最多更新n次  做了…

Two Rabbits Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 505    Accepted Submission(s): 260 Problem Description Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny…