题目链接:http://codeforces.com/contest/727/problem/F 题目大意:有n个问题,每个问题有一个价值ai,一开始的心情值为q,每当读到一个问题时,心情值将会加上该问题的价值.问题只能按顺序读.有m个询问,求当q=bi时,至少要删去多少个问题才能使得在任何时候心情值都>=0 参考这篇:http://blog.csdn.net/aufeas/article/details/53031439 解法一: 贪心 分别求出最多删去i个问题需要的初始心情值的最小值f[i]…

Polycarp's problems time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Polycarp is an experienced participant in Codehorses programming contests. Now he wants to become a problemsetter. He se…

链接 Codeforces 659F Polycarp and Hay 题意 一个矩阵,减小一些数字的大小使得构成一个连通块的和恰好等于k,要求连通块中至少保持一个不变 思路 将数值从小到大排序,按顺序把与其相邻的加到并查集中.记录当前并查集中的个数,如果当前值能被K整除且总和超过了K,那么就可以以该点为中心输出了. 代码 #include <iostream> #include <cstdio> #include <vector> #include <stack…

Description 有一个长度为 \(n\) 有正负权值的序列,你一开始有一个值,每次到一个权值就加上,最少需要删掉多少数值才能到序列末尾.\(n \leqslant 750,m \leqslant 2 \times 10^5\) Sol DP+二分. 发现这个东西有后效性,就是前面选不选会影响后面的决策,并且权值太大无法记录. 但是我们可以倒着做,因为后面的决策无法影响前面的决策. \(f[i][j]\) 表示到 \(i\) 删掉 \(j\) 个至少需要的初始权值. 因为初始权值非负,所以…

C. Polycarp at the Radio time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented a…

Description Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a1, a2, ..., an, where ai is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to…

Description Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters. Let A be a set of positions in the string. Let's call it pretty if following conditions are met…

913D - Too Easy Problems 思路:二分check k 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) ; struct pro{ int t,a,id; bool operator < (pro tmp){ return t<tmp.t; } }a[N…

861D - Polycarp's phone book 思路:用map做的话,只能出现一次循环,否则会超时. 代码: #include<bits/stdc++.h> using namespace std; #define f first #define s second #define pb push_back #define mp make_pair map<string,int> ma; ]; int main(){ int n; string s,nu; cin>&…

D. Polycarp's Picture Gallery time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Polycarp loves not only to take pictures, but also to show his photos to friends. On his personal website he h…