E. Thief in a Shop 题目连接: http://www.codeforces.com/contest/632/problem/E Description A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an inf…

E - Thief in a Shop 题目大意:给你n ( n <= 1000)个物品每个物品的价值为ai (ai <= 1000),你只能恰好取k个物品,问你能组成哪些价值. 思路:我们很容易能够想到dp[ i ][ j ]表示取i次j是否存在,但是复杂度1e12肯定不行. 我们将ai排序,每个值都减去a[1]然后再用dp[ i ]表示到达i这个值最少需要取几次,只需要1e9就能完成, 我们扫一遍dp数组,如果dp[ i ]  <= k 则说明 i + k * a[1]是能取到的.…

E. Thief in a Shop   A thief made his way to a shop. As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product o…

Educational Codeforces Round 63 (Rated for Div. 2) D. Beautiful Array time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an array aa consisting of nn integers. Beauty of array i…

https://codeforces.com/contest/1101/problem/F 题意 有n个城市,m辆卡车,每辆卡车有起点\(s_i\),终点\(f_i\),每公里油耗\(c_i\),可加油次数\(r_i\),每辆卡车的油箱大小一样,每次加油都会加满,在城市之间不能加油,问最小油箱大小能满足每辆卡车顺利到达终点 题解 n<=400,m<=250000,考虑离线处理出任意两个城市能加油k次的最小油耗,然后对于每辆卡车询问 定义dp[l][r][k]为区间[l,r]能分成(k+1)段各…

https://codeforces.com/contest/1155/problem/D 题意 一个n个数的数组\(a[i]\),可以选择连续的一段乘x,求最大连续子序列的值 题解 错误思路:贪心,假如x<0,那么就选择最小的一段乘以x,再求最大连续子序列,因为这一段可能夹着一些很大的正数使得翻转一整段的代价很大,可能单独翻转前半段或者后半段更好 定义\(f[i]\)为以i结尾的最大连续子序列,\(g[i]\)为以i开头的最大连续子序列 假设选择翻转的区间是(l,r),则有\(ans=max(…

https://codeforces.com/contest/1096/problem/D 题意 给一个串s,删掉一个字符的代价为a[i],问使得s的子串不含"hard"的最小代价 题解 定义\(dp[i][j]\)为到第i位下一个将要匹配j的最小代价 \(若s[i]==t[j]\) 删掉:\(min(dp[i+1][j],dp[i][j]+a[i])\) 不删,若j<3:\(min(dp[i+1][j+1],dp[i][j])\) \(若s[i]!=t[j]\) 不用删:\(m…

Generate a String 题目链接: http://codeforces.com/contest/710/problem/E Description zscoder wants to generate an input file for some programming competition problem. His input is a string consisting of n letters 'a'. He is too lazy to write a generator s…

题:https://codeforces.com/contest/1257/problem/E 题意:给定3个数组,可行操作:每个数都可以跳到另外俩个数组中去,实行多步操作后使三个数组拼接起来形成升序. 输出最小操作次数 dp: #include<bits/stdc++.h> using namespace std; ; ]; int a[M]; ///dp[i][0]表示前i个数全属于第一个数组所要花费的最小代价 ///dp[i][1]表示前i个数全属于第一.二个数组所要花费的最小代价 //…

题目链接: 题目 E. Another Sith Tournament time limit per test2.5 seconds memory limit per test256 megabytes inputstandard input outputstandard output 问题描述 The rules of Sith Tournament are well known to everyone. n Sith take part in the Tournament. The Tour…