Codeforces Gym100735 G.LCS Revised (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)
G.LCS Revised
The longest common subsequence is a well known DP problem: given two strings A and B, one has to compute the maximum length of a subsequence that's common to both A and B.
In this particular problem we work with strings A and B formed only by 0 and 1, having the same length. You're given a string A of length n. Iterate all strings B possible. There are 2n of them. Calculate, for each string B, the longest common subsequence of A and B. Then, output the minimum length obtained.
Input
The first and the only line of the input contains string A, formed only by 0 and 1. It's guaranteed that the length is between 1 and 105.
Output
Output a single number - the requested length.
Example
101010
3
超级无敌大水题,直接找0和1哪个少就可以。。。
代码:
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 using namespace std;
5 const int N=1e6+10;
6 char a[N];
7 int main(){
8 while(~scanf("%s",a)){
9 int len=strlen(a);
10 int ans,num1=0,num2=0;
11 for(int i=0;i<len;i++){
12 if(a[i]=='1')num1++;
13 else num2++;
14 }
15 ans=min(num1,num2);
16 printf("%d\n",ans);
17 }
18 return 0;
19 }
Codeforces Gym100735 G.LCS Revised (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)的更多相关文章
- Codeforces Gym100735 D.Triangle Formation (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)
日常训练题解 D.Triangle Formation You are given N wooden sticks. Your task is to determine how many triang ...
- Codeforces Gym100735 I.Yet another A + B-Java大数 (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)
I.Yet another A + B You are given three numbers. Is there a way to replace variables A, B and C with ...
- Codeforces Gym100735 E.Restore (KTU Programming Camp (Day 1) Lithuania, Birˇstonas, August 19, 2015)
E - Restore Given a matrix A of size N * N. The rows are numbered from 0 to N-1, the columns are num ...
- 【KTU Programming Camp (Day 3)】Queries
http://codeforces.com/gym/100739/problem/A 按位考虑,每一位建一个线段树. 求出前缀xor和,对前缀xor和建线段树. 线段树上维护区间内的0的个数和1的个数 ...
- KTU Programming Camp (Winter Training Day 1)
A.B.C(By musashiheart) 0216个人赛前三道题解 E(By ggg) Gym - 100735E Restore H(by pipixia) Gym - 100735H
- CodeForcesGym 100735G LCS Revised
LCS Revised Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on CodeForcesGym. O ...
- [codeforces 549]G. Happy Line
[codeforces 549]G. Happy Line 试题描述 Do you like summer? Residents of Berland do. They especially love ...
- CodeForces 794 G.Replace All
CodeForces 794 G.Replace All 解题思路 首先如果字符串 \(A, B\) 没有匹配,那么二元组 \((S, T)\) 合法的一个必要条件是存在正整数对 \((x,y)\), ...
- Codeforces 1207 G. Indie Album
Codeforces 1207 G. Indie Album 解题思路 离线下来用SAM或者AC自动机就是一个单点加子树求和,套个树状数组就好了,因为这个题广义SAM不能存在 \(len[u] = l ...
随机推荐
- The Moving Points - HDU - 4717 (模拟退火)
题意 二维空间中有\(n\)个运动的点,每个点有一个初始坐标和速度向量.求出一个时间\(T\),使得此时任意两点之间的最大距离最小.输出\(T\)和最大距离. 题解 模拟退火. 这个题告诉了我,初始步 ...
- spark 对hbase 操作
本文将分两部分介绍,第一部分讲解使用 HBase 新版 API 进行 CRUD 基本操作:第二部分讲解如何将 Spark 内的 RDDs 写入 HBase 的表中,反之,HBase 中的表又是如何以 ...
- HDU 3333 Turing Tree 莫队算法
题意: 给出一个序列和若干次询问,每次询问一个子序列去重后的所有元素之和. 分析: 先将序列离散化,然后离线处理所有询问. 用莫队算法维护每个数出现的次数,就可以一边移动区间一边维护不同元素之和. # ...
- TCP报文格式,TCP的三次握手和四次挥手&hosts文件
1.TCP报文格式 TCP报头中的源端口号和目的端口号同IP数据报中的源IP与目的IP唯一确定一条TCP连接 序号(4字节=32位): 37 59 56 75 用来标识TCP发端向TCP收端发送的数据 ...
- MOCTF 简单注入
最近在练习sql注入写脚本,记录一下思路,刚学的and 1=1也拿出来溜溜 http://119.23.73.3:5004/?id=1 首先,没有被过滤是正常显示. 没有被过滤但是查询不到就是空白,比 ...
- 微信小程序-----校园头条详细开发之列表展示数据
1.分类列表数据展示功能的实现 1.1 结构 1.2 代码实现 1.2.1 列表显示数据,.每次界面显示6条数据,发请求获取数据,动态存放 var app = getApp() Page({ dat ...
- maven学习(十一)——maven中的聚合与继承
一.聚合 如果我们想一次构建多个项目模块,那我们就需要对多个项目模块进行聚合 1.1.聚合配置代码 <modules> <module>模块一</module> & ...
- idea使用maven逆向mybitis的文件
引用自 http://blog.csdn.net/for_my_life/article/details/51228098 本文介绍一下用Maven工具如何生成Mybatis的代码及映射的文件. 一. ...
- 构建乘积数组--java
题目:给定一个数组A[0,1,...,n-1],请构建一个数组B[0,1,...,n-1],其中B中的元素B[i]=A[0]*A[1]*...*A[i-1]*A[i+1]*...*A[n-1].不能使 ...
- 被readLine()折腾了一把
虽然写IO方面的程序不多,但BufferedReader/BufferedInputStream倒是用过好几次的,原因是: 它有一个很特别的方法:readLine(),使用起来特别方便,每次读回来的都 ...