题目链接:http://codeforces.com/problemset/problem/752/C time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Santa Claus has Robot which lives on the infinite grid and can move along its lines. He…
Problem Codeforces Global Round 1 - D. Jongmah Time Limit: 3000 mSec Problem Description Input Output Print one integer: the maximum number of triples you can form. Sample Input 10 62 3 3 3 4 4 4 5 5 6 Sample Output 3 题解:动态规划,对这种状态定义不熟悉,主要还是没有发现最优方…
Problem Codeforces #541 (Div2) - E. String Multiplication Time Limit: 2000 mSec Problem Description Input Output Print exactly one integer — the beauty of the product of the strings. Sample Input 3aba Sample Output 3 题解:这个题的思维难度其实不大,需要维护什么东西很容易想到,或…
B. z-sort time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold: ai ≥ …
D. Minimum Triangulation time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a regular polygon with nn vertices labeled from 11 to nn in counter-clockwise order. The triangulatio…
题目链接:https://codeforces.com/contest/1090/problem/D Vasya had an array of n integers, each element of the array was from 1 to n. He chose m pairs of different positions and wrote them down to a sheet of paper. Then Vasya compared the elements at these…
C. Maximal GCD time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a1, a2, ...…
[题目]C. Bear and Company [题意]给定大写字母字符串,交换相邻字符代价为1,求最小代价使得字符串不含"VK"子串.n<=75. [算法]动态规划 [题解]关键在于表示状态,我们将确定下来的前若干个固定作为状态,后面新加的字符不会进入固定的前若干个.(为了方便,非'V''K'的字符皆为‘X') 由于相同字符显然不可能跨越,那么前若干个的有效信息只有:它是由前v个’V',前k个‘K',前x个’X'组成的,最后一个字符是否’V',即f[v][k][x][0/1].…
题意 给定一个长度为n的二进制串(即由n个'0'和'1'构成的字符串),你最多可以进行k次交换相邻两个字符的操作,求字典序最小的串. 思路 大致就是找0的位置,然后贪心的放到最前面,这样字典序会最小: 代码 我的丑做法: #include<bits/stdc++.h> using namespace std; #define inf 0x3f3f3f3f #define ll long long const int N=200005; const int mod=1e9+7; const do…