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乱搞

2017

codeforces edu #29 E. Turn Off The TV (思维,乱搞)

·737 words·2 mins
题目链接 题意:有若干线段,给出起点和终点,问是否有一个线段是冗余的。冗余的意思是说,对于该线段所覆盖的所有整数点,没有该线段,也能被其他一个或者多个线段覆盖到。如果有,输出任意一个冗余线段即可。

2016

BZOJ 1637: [Usaco2007 Mar]Balanced Lineup (前缀和乱搞)

·1332 words·3 mins
1637: [Usaco2007 Mar]Balanced Lineup # Time Limit: 5 Sec Memory Limit: 64 MB Submit: 503 Solved: 336 [Submit][Status][Discuss] Description # Farmer John 决定给他的奶牛们照一张合影,他让 N (1 ≤ N ≤ 50,000) 头奶牛站成一条直线,每头牛都有它的坐标(范围: 0..1,000,000,000)和种族(0或1)。 一直以来 Farmer John 总是喜欢做一些非凡的事,当然这次照相也不例外。他只给一部分牛照相,并且这一组牛的阵容必须是“平衡的”。平衡的阵容,指的是在一组牛中,种族0和种族1的牛的数量相等。 请算出最广阔的区间,使这个区间内的牛阵容平衡。区间的大小为区间内最右边的牛的坐标减去最左边的牛的坐标。 输入中,每个种族至少有一头牛,没有两头牛的坐标相同。

codeforces #346 div 2 C. Tanya and Toys (暴力乱搞)

·613 words·2 mins
题目链接 题意:有1E9个礼物,第i个礼物价钱是i,然后现在已经有n个不重复的礼物,a[i],m元钱,想尽可能多得买不同种类的礼物,还能买多少个。 思路:先不考虑已经买的,从1连续买到k,然后考虑在这个区间内已经买的,等于实际上没有花钱。 反正就是暴力搞啊搞啊。。我也不知道怎么搞。。 结果最后。。方案数为0的时候。。。最后一个答案我是单独输出的。。忘了判断了。。。所以会输出两个0.宝宝心里苦啊。。。。。。。。

codeforces 484 B Maximum Value (暴力乱搞)

·504 words·2 mins
题目链接 题意:给出n个元素的序列,求出最大的a[i]%a[j] (i>=j) 思路:没思路。。。。 Let us iterate over all different a__j. Since we need to maximize , then iterate all integer x (such_x_ divisible by a__j) in range from 2_a__j_ to M, where M — doubled maximum value of the sequence. For each such x we need to find maximum a__i, such a__i < x. Limits for numbers allow to do this in time O(1) with an array. After that, update answer by value . Total time complexity is O(nlogn + MlogM)

2015

codeforces #322 div 2 C. Developing Skills(乱搞)

·1107 words·3 mins
time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Petya loves computer games. Finally a game that he’s been waiting for so long came out! The main character of this game has n different skills, each of which is characterized by an integer a__i from 0 to 100. The higher the number_a__i_ is, the higher is the i-th skill of the character. The total rating of the character is calculated as the sum of the values of for all i from 1 to n. The expression ⌊ x⌋ denotes the result of rounding the number x down to the nearest integer.