Codeforce466A——贪心——DZY Loves Sequences

http://codeforces.com/problemset/problem/446/A

/*
贪心,开前缀后缀,枚举,每个i
*/
/************************************************
* Author        :Powatr
* Created Time  :2015-8-16 20:03:03
* File Name     :cf466A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[MAXN];
int pre1[MAXN], pre2[MAXN];
int main(){
    int n;
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++)
            pre1[i] = pre2[i] = 1;
        for(int i = 2; i <= n; i++){
            if(a[i] > a[i-1])
                pre1[i] = pre1[i-1] + 1;
        }
        for(int i = n - 1; i >= 1; i--){
            if(a[i+1] > a[i]){
                pre2[i] = pre2[i+1] + 1;
            }
        }
        int ans = 1;
        if(n > 1) ans = max(pre1[n-1] + 1, pre2[2] + 1);
            for(int i = 2; i <= n - 1; i++){
                if(a[i+1] - a[i-1] >= 2){
                    ans = max(ans, pre1[i-1] + pre2[i+1] + 1);
                }
                else {
                    ans = max(ans , pre1[i-1] + 1);
                    ans = max(ans, pre2[i+1] + 1);
                }
        }
        printf("%d\n", ans);
    }
    return 0;
}

  

时间: 2024-11-12 18:19:53

Codeforce466A——贪心——DZY Loves Sequences的相关文章

Codeforces Round #FF(255) C. DZY Loves Sequences (LIS升级)

题目:C. DZY Loves Sequences (LIS升级) 题意: 在n个数中,最多改变一个数字,并求能够达到的最长严格上升子序列(连续)长度 分析: 考虑第i个数,能否改变后拼接前后两个字串,并维护当前最大值 状态: left[i]:  表示以i为终点的最长严格上升子序列长度 right[i]: 表示以i为起点的最长严格上升子序列长度 dp[i]:   表示改变第i个数后,拼接前后字串的长度 转移方程:       dp[i] = max{left[i-1] + right[i+1] 

[2016-04-13][codeforces][447][C][DZY Loves Sequences]

时间:2016-04-13 23:39:47 星期三 题目编号:[2016-04-13][codeforces][447][C][DZY Loves Sequences] 题目大意:给定一串数字,问改变其中一个数字之和,最长能得到多长的严格增加的子串 分析: 维护每个数字往左和往右能延续多长(严格减,增),然后枚举每个点, 如果这个点已经在一个严格增加的序列中,那么ans =min(n, max(ans , l[i] + r[i] + 1)) 即左右两边延伸之后,改变后面非递增的一个数字 注意这

Codeforces Round #FF(255) DIV2 C - DZY Loves Sequences

A - DZY Loves Hash 水题,开辟一个数组即可 #include <iostream> #include <vector> #include <algorithm> #include <string> using namespace std; int main(){ int p,n; cin >> p >> n; vector<bool> buckets(302,false); bool flag = fal

DP Codeforces Round #FF (Div. 1) A. DZY Loves Sequences

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 MAXN =

codeforces#FF(div2) DZY Loves Sequences

n个数,可以任意改变其中一个数,求最长的上升子区间长度 思路:记录一个from[i]表示从位置i的数开始最长的上升区间长度 记录一个to[i]表示到位置i的数所能达到的最长上升区间长度 枚举要改变的数的位置i,此时能达到的长度为to[i - 1] + from[i + 1] + 1,取最大值 //#pragma comment(linker, "/STACK:102400000,102400000") //HEAD #include <cstdio> #include &l

[CodeForces - 447C] C - DZY Loves Sequences

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

codeforces 446A DZY Loves Sequences

codeforces   446A   DZY Loves Sequences         题目链接:http://codeforces.com/problemset/problem/446/A 题目大意:给出一个定长为n的数列a,问改动当中一个数后.可能出现的最长严格上升子段的长度是多少. 题目分析:先不考虑"改动当中一个数"这个条件,这样问题就简单多了,从前到后遍历计数就可以(定义一个数组inc[]长度同a,初始化全部点为1,遍历假设当前点a[i]>a[i-1]就置inc

Codeforces 447C - DZY Loves Sequences

447C - DZY Loves Sequences 思路:dp 代码: #include<bits/stdc++.h> using namespace std; #define ll long long const int INF=0x3f3f3f3f; const int N=1e5+5; int a[N]; int f[N],g[N]; int main() { ios::sync_with_stdio(false); cin.tie(0); int n; cin>>n; f

Codeforces Round #FF (Div. 2) C - DZY Loves Sequences (DP)

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 longest subsegment of a, su