E1. String Coloring (easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output This is an easy version of the problem. The actual problems are different, but the easy version is almost…

题意:有\(n\)本书,A和B都至少要从喜欢的书里面读\(k\)本书,如果一本书两人都喜欢的话,那么他们就可以一起读来节省时间,问最少多长时间两人都能够读完\(k\)本书. 题解:我们可以分\(3\)种情况来存,即: ​ 1.\(a=b=1\). 2.\(a=1,b=0\). 3.\(a=0,b=1\). 对于2和3来说,我们可以将他们排序,然后合并到一起,最后放到第1种情况中再排一次序,取前\(k\)个前缀和即可. 代码: int n,k; int t,x,y; int ans; vector…

题意: 给出一个由n个数字组成的数组,先让你找出符合下列条件的子集的数量: 每个子集包含的数字个数为m = 3 这三个数字中的最大值减去最小值不超过k = 2 思路: 首先对给出的数组进行排序,现在假设这个数组为\(a\),这个子集为\(\{A_1, A_2, A_3\}\),那么我们每次枚举\(A_1\),用一个指针记录数组中最后一个满足\(a[p] - A_1\)的位置,那么如果\(p\)和\(A_1\)之间元素的个数n再加上1大于等于2,那么答案的总数就加上\(C_n^2\). 这里有个地…

http://codeforces.com/problemset/problem/1216/E1 E1. Numerical Sequence (easy version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between the easy and the hard ver…

B. Ping-Pong (Easy Version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to in…

3868 - Earthstone: Easy Version Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3867 Description Earthstone is a famous online card game created by Lizard Entertainment. It is a collectible card g…

题目链接:Pictures with Kittens (easy version) 题意:给定n长度的数字序列ai,求从中选出x个满足任意k长度区间都至少有一个被选到的最大和. 题解:$dp[i][j]$:以i为结尾选择j个数字的最大和. $dp[i][j]=max(dp[i][j],dp[s][j-1]+a[i])$,$s为区间[i-k,i)$. 以i为结尾的最大和可以由i之前k个位置中的其中一个位置选择j-1个,再加上当前位置的ai得到. #include <cstdio> #includ…

Problem UVA12569-Planning mobile robot on Tree (EASY Version) Accept:138  Submit:686 Time Limit: 3000 mSec  Problem Description  Input The first line contains the number of test cases T (T ≤ 340). Each test case begins with four integers n, m, s, t (…

Coffee and Coursework (Easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions is the constraints. Polycarp has to write a coursewor…

1114 ModricWang's FFT EASY VERSION 思路 利用FFT做大整数乘法,实际上是把大整数变成多项式,然后做多项式乘法. 例如,对于\(1234\),改写成\(f(x)=1*x^3+2*x^2+3*x+4\),那么\(x=10\)处的值就是原数.类似的,对于输入的两个大整数,转换为\(f(x)\) 和\(g(x)\) ,利用FFT求出\(h(x)=f(x)*g(x)\) ,此时\(h(10)\) 就是乘积. 代码 #include <cstdio> #include…