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…
Problem Codeforces #541 (Div2) - F. Asya And Kittens Time Limit: 2000 mSec Problem Description Input The first line contains a single integer nn (2≤n≤150000) — the number of kittens. Each of the following n−1lines contains integers xi and yi (1≤xi,…
Problem Codeforces #541 (Div2) - D. Gourmet choice Time Limit: 2000 mSec Problem Description Input Output The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwis…
Problem Codeforces #548 (Div2) - D.Steps to One Time Limit: 2000 mSec Problem Description Input The first and only line contains a single integer mm (1≤m≤100000,1≤m≤100000). Output Print a single integer — the expected length of the array aa writte…
# [Codeforces #312 div2 A]Lala Land and Apple Trees 首先,此题的大意是在一条坐标轴上,有\(n\)个点,每个点的权值为\(a_{i}\),第一次从原点开始走,方向自选(<- or ->),在过程中,若遇到一个权值>0的点,则将此权值计入答案,并归零.当次.此方向上的所有点均为0后,输出此时的答案. 然后,进行分析: 我们很容易想到这是一个贪心,我们将正的和负的分别存入两个数组,最初的方向为: \(zhengsum > fusum…
题目链接:D Persistent Bookcase 题意:有一个n*m的书架,开始是空的,现在有k种操作: 1 x y 这个位置如果没书,放书. 2 x y 这个位置如果有书,拿走. 3 x 反转这一行,即有书的位置拿走,没书的位置放上书. 4 x 返回到第x步操作之后的书架. 现在给出q个操作,询问每次操作之后书架上书的数量. 思路: 开始没有思路.后来被告知dfs. [词不达意.参考:http://blog.csdn.net/zyjhtutu/article/details/5227949…
E. Subordinates time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n workers in a company, each of them has a unique id from 1 to n. Exaclty one of them is a chief, his id is s. Each…
题目链接:http://codeforces.com/contest/699/problem/C dp[i][j]表示第i天做事情j所得到最小的假期,j=0,1,2. #include<bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f; int dp[105][3]; int main() { int n; scanf("%d",&n); memset(dp,INF,sizeof(dp)); d…
题目链接:http://codeforces.com/contest/706/problem/C #include<bits/stdc++.h> using namespace std; typedef long long ll; const int N=1e5+3; const ll INF=1e18; ll dp[N][2]; string a[N],b[N]; int c[N]; int main() { int n; scanf("%d",&n); for(…
