D2. Kirk and a Binary String (hard version) time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the length of the string. You can hack this pro…

题意:https://codeforc.es/problemset/problem/1204/D2 给你一个01串,如:0111001100111011101000,让你改这个串(使0尽可能多,任意 l~r 的LIS最长上升子序列长度不变). 问:0111001100111011101000 答:0011001100001011101000 思路: 参考题解:考虑什么串无法该变. 1. 10无法被改变(我们称为固定串),00,01,11都可以改. 2. 固定串+固定串(无法被改变). 3. 1+…

D1. Kirk and a Binary String (easy version) 01串找最长不降子序列 给定字符串s,要求生成一个等长字符串t,使得任意l到r位置的最长不降子序列长度一致 从后往前暴力枚举,枚举每个一替换成0后是否改变了l到r位置的最长不降子序列长度 01串的最长不降子序列,可以通过线性dp求解 dp i表示以i结尾的最长不降子序列长度 dp[0]=dp[0]+s[i]=='0'; dp[1]=max(dp[0],dp[1])+s[i]=='1'; #include<bi…

#define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;char s[100007];int main(){ cin>>s+1; int n=strlen(s+1); int cnt=0; for(int i=n;i>=1;--i){//从后向前,保证后面的解都是合法的情况下 if(s[i]=='1'){//如果当前位置的数字是1 if(cnt)//i后面1的个数小于0的个数,此时如果把i位…

题目链接:http://codeforces.com/contest/1204/problem/D2 题目是给定一个01字符串,让你尽可能多地改变1变为0,但是要保证新的字符串,对任意的L,R使得Sl,Sl+1,Sl+2...Sr的最长不递减子序列长度保持不变,求新的串s. dp思路,从前往后遍历串s. 1 . 遇到s[ i ] = 0 是不能改变的,因为从i到n的最长不递减子序列必定是以s[ i ] = 0为起点的,改变之后会减少长度. 2 . 遇到s[ i ] = 1.我们考虑如果变为0,首…

A. BowWow and the Timetable time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output In the city of Saint Petersburg, a day lasts for 2100 minutes. From the main station of Saint Petersburg, a train dep…

Problem   Codeforces Round #539 (Div. 2) - D. Sasha and One More Name Time Limit: 1000 mSec Problem Description Input The first line contains one string s (1≤|s|≤5000) — the initial name, which consists only of lowercase Latin letters. It is guarante…

题目传送门 /* 题意:套娃娃,可以套一个单独的娃娃,或者把最后面的娃娃取出,最后使得0-1-2-...-(n-1),问最少要几步 贪心/思维题:娃娃的状态:取出+套上(2),套上(1), 已套上(0),先从1开始找到已经套好的娃娃层数, 其他是2次操作,还要减去k-1个娃娃是只要套上就可以 详细解释:http://blog.csdn.net/firstlucker/article/details/46671251 */ #include <cstdio> #include <algor…

题目传送门 /* 题意:选择k个m长的区间,使得总和最大 01背包:dp[i][j] 表示在i的位置选或不选[i-m+1, i]这个区间,当它是第j个区间. 01背包思想,状态转移方程:dp[i][j] = max (dp[i-1][j], dp[i-m][j-1] + sum[i] - sum[i-m]); 在两个for循环,每一次dp[i][j]的值都要更新 */ #include <cstdio> #include <cstring> #include <algorit…

Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a descr…