题目地址:http://codeforces.com/contest/447/problem/C C. DZY Loves Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output DZY has a sequence a, consisting of n integers. We'll call a sequence…

题目:click here 题意:求给定序列更改其中一个元素后的最长连续上升子序列的长度 分析:最长的连续子序列有2种,一种是严格上升(没有更改元素)的长度加1,一种是两段严格上升的加起来. #include <bits/stdc++.h> using namespace std; #define F first #define S second #define pb push_back #define power(a) ((a)*(a)) #define ENTER printf("…

题意:给定一个序列,让你找出一个最长的序列,使得最多改其中的一个数,使其变成严格上升序列. 析:f[i] 表示以 i 结尾的最长上升长度,g[i] 表示以 i 为开始的最长上升长度,这两个很容易就求得,最后枚举中间值即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib>…

首先要知道选择行列操作时顺序是无关的 用两个数组row[i],col[j]分别表示仅选择i行能得到的最大值和仅选择j列能得到的最大值 这个用优先队列维护,没选择一行(列)后将这行(列)的和减去对应的np (mp)又一次增加队列 枚举选择行的次数为i,那么选择列的次数为k - i次,ans = row[i] + col[k - i] - (k - i) * i * p; 既然顺序无关,能够看做先选择完i次行,那么每次选择一列时都要减去i * p,选择k - i次列,即减去(k - i) * i *…

题目传送门 /* DP:先用l,r数组记录前缀后缀上升长度,最大值会在三种情况中产生: 1. a[i-1] + 1 < a[i+1],可以改a[i],那么值为l[i-1] + r[i+1] + 1 2. l[i-1] + 1 3. r[i+1] + 1 //修改a[i] */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF…

预处理出每一个数字能够向后延伸多少,然后尝试将两段拼起来. C. DZY Loves Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output DZY has a sequence a, consisting of n integers. We'll call a sequence ai, ai + 1, ..., a…

A. DZY Loves Sequences 题目连接: http://www.codeforces.com/contest/446/problem/A Description DZY has a sequence a, consisting of n integers. We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) d…

C. DZY Loves Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output DZY has a sequence a, consisting of n integers. We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment…

C - DZY Loves Sequences DZY has a sequence a, consisting of n integers. We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment. Your task is to find the lon…

447C - DZY Loves Sequences 思路:dp 代码: #include<bits/stdc++.h> using namespace std; #define ll long long const int INF=0x3f3f3f3f; ; int a[N]; int f[N],g[N]; int main() { ios::sync_with_stdio(false); cin.tie(); int n; cin>>n; f[]=; cin>>a[…