题意:给你两个字符串,每次取它们的子串C和D,然后求LCS,得到的贡献为\(4*LCS(C,D)-|C|-|D|\),求最大贡献. 题解:首先应该了解\(O(n^2)\)的LCS的dp写法,然后在此基础上稍加改动,对于子串\(C\)和\(D\),如果\(c[i]=d[j]\),那么他们的LCS应该\(+1\),长度也分别\(+1\),所以\(dp[i][j]=dp[i-1][j-1]+2\).而如果\(c[i]\)和\(d[j]\)不匹配,因为长度加一,所以贡献应该减一,但是由最大子段和,我们应…

比赛链接:https://codeforces.com/contest/1447 A. Add Candies 题意 \(1\) 到 \(n\) 个袋子里依次有 \(1\) 到 \(n\) 个糖果,可以选择操作 \(m\) 次,第 \(i\) 次操作可以: 选择一个袋子,除了这个袋子外其他袋子都再放入 \(i\) 个糖果 问是否存在一种操作序列,使得最终 \(n\) 个袋子内的糖果数相同. 题解 将每个袋子内的糖果都加到 \(\frac{n(n + 1)}{2}\) 个. 代码 #include…

A 初始情况\(1\) ~ \(n\)堆分别有 \(1\) ~ \(n\) 个糖果,第\(i\)次操作给除了所选堆的糖果数 \(+ i\), 找到一种方案可以使得所有堆糖果数相同,输出操作次数和每次选定不变的堆. 选定当前堆不变,其他堆\(+ i\)等价于将选定堆\(-i\), 故只需要考虑如何将所有堆减到\(0\)即可, 即操作\(n\)次,第\(i\)次选择第\(i\)堆. #include <bits/stdc++.h> using namespace std; int main() {…

A. Mahmoud and Longest Uncommon Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest comm…

A. Knapsack 猜个结论--先把所有的东西加起来,如果小于 \(\frac{1}{2}m\) 就输出不合法:如果在 \([\frac{1}{2}m, m]\)之间直接全部输出:若大于 \(m\),那就想办法把他减到 \(m\) 以下并且大于等于 \(\frac{1}{2}m\),那么问题就转化为了求序列减完以后大于等于 \(\frac{1}{2}m\) 的情况下的最小值.那我们排个序,从大到小循环,把当前能减的都减掉就是了. for (int i = n; i; i--) { //fou…

题目链接 http://codeforces.com/contest/711/problem/C Description ZS the Coder and Chris the Baboon has arrived at Udayland! They walked in the park where n trees grow. They decided to be naughty and color the trees in the park. The trees are numbered wit…

E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not e…

D - Chloe and pleasant prizes 链接 http://codeforces.com/contest/743/problem/D 题面 Generous sponsors of the olympiad in which Chloe and Vladik took part allowed all the participants to choose a prize for them on their own. Christmas is coming, so sponso…

题目链接: http://www.codeforces.com/contest/675/problem/E 题意: 对于第i个站,它与i+1到a[i]的站有路相连,先在求所有站点i到站点j的最短距离之和(1<=i<j<=n) 题解: 这种所有可能都算一遍就会爆的题目,有可能是可以转化为去求每个数对最后答案的贡献,用一些组合计数的方法一般就可以很快算出来. 这里用dp[i]表示第i个站点到后面的所有站的最短距离之和,明显,i+1到a[i]的站,一步就可以到,对于后面的那些点,贪心一下,一定…

题目链接:http://codeforces.com/contest/560/problem/E 给你一个n*m的网格,有k个坏点,问你从(1,1)到(n,m)不经过坏点有多少条路径. 先把这些坏点排序一下. dp[i]表示从(1,1)到第i个坏点且不经过其他坏点的路径数目. dp[i] = Lucas(x[i], y[i]) - sum(dp[j]*Lucas(x[i]-x[j], y[i]-x[j])) , x[j] <= x[i] && y[j] <= y[i] //到i…