Educational Codeforces Round 39 G Almost Increasing Array

传送门

非减子序列普通lcs Dp TLE ,换成upper_bound 竟然过了，神奇

#include <bits/stdc++.h>
using namespace std;
typedef long long ll ;
typedef double    dl ;
#define INF ((ll)2e9)const int  inf = 987654321;
const int sz = 1e6 + 5;
const int mod = 1e9 + 7;
const int sqrtn = 300;#define f(i,l,r) for(int i=l;i<=r;++i)
#define g(i,l,r) for(int i=l;i>=r;--i)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define sz(a)   strlen(a)
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);#define lowbit(x) x&(-x)
#define X first
#define Y second #define pb push_back
#define pii pair<int,int>int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};void work1()
{int seq1[] = {1, 2, 3, 3, 4, 5}, seq2[] = {9, 8, 7, 7, 6, 5};cout<<upper_bound(seq1, seq1+6, 3) - seq1<<endl;cout<<lower_bound(seq1, seq1+6, 3) - seq1<<endl;cout<<endl;cout<<upper_bound(seq2, seq2+6, 7, greater<int>()) - seq2<<endl;cout<<lower_bound(seq2, seq2+6, 7, greater<int>()) - seq2<<endl;
}
ll n,dp[2][sz];
ll ans,x;
void work()
{cin>>n;f(i,0,sz-1)dp[0][i]=dp[1][i]=INF;dp[0][0]=-INF;ll change =0;f(i,1,n){cin>>x;ll id = upper_bound(dp[1],dp[1]+sz,x-i+1)-dp[1];dp[1][id]=x-i+1;ans=max(ans,id);dp[1][change]=min(dp[0][change],dp[1][change]);ans=max(ans,change);id = upper_bound(dp[0],dp[0]+sz,x-i)-dp[0];change=id;dp[0][id]=x-i; }cout<<n-ans-1<<"\n";
}int main()
{FAST_IO ;freopen("in","r",stdin);freopen("out","w",stdout);cout<<"hello"<<endl;work3();return 0;
}

转载于:https://www.cnblogs.com/corx/p/8523892.html

Educational Codeforces Round 39 G Almost Increasing Array相关推荐

Educational Codeforces Round 25 G. Tree Queries
题目链接:Educational Codeforces Round 25 G. Tree Queries 题意: 给你一棵树,一开始所有的点全是黑色,有两种操作. 1 x 将x这个点变为黑色,保证第一 ...
Educational Codeforces Round 32 G. Xor-MST 01tire + 分治 + Boruvka
传送门文章目录题意: 思路: 题意: 给你一个长度为nnn序列aaa,每两个点之间的边权为ai⊕aja_i\oplus a_jai⊕aj,问你最小生成树的权值是多少. n≤2e5,ai< ...
Educational Codeforces Round 37 G. List Of Integers （二分，容斥定律，数论）
G. List Of Integers time limit per test 5 seconds memory limit per test 256 megabytes input standard ...
Educational Codeforces Round 39 B Weird Subtraction Process
传送门模拟,写的比较土,别吐槽 #include <bits/stdc++.h> using namespace std; typedef long long ll ; typedef ...
Educational Codeforces Round 39 F Largest Beautiful Number
传送门暴力DP dp[i][j][k] 从j 到 k 发f(x) 的大小, #include <bits/stdc++.h> using namespace std; typedef l ...
Educational Codeforces Round 39 A Partition
传送门分治就好了,哈哈大于小于0的数据分开算 #include <bits/stdc++.h> using namespace std; typedef long long ll ; t ...
Educational Codeforces Round 21 G Anthem of Berland
题目链接 Berland has a long and glorious history. To increase awareness about it among younger citizens, ...
Educational Codeforces Round 106 (Rated for Div. 2)（A ~ E）题解（每日训练 Day.16 ）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划目录 Educational Codeforces Round 106 (Rated for Div. ...
Educational Codeforces Round 32
http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 ...

Educational Codeforces Round 39 G Almost Increasing Array

Educational Codeforces Round 39 G Almost Increasing Array相关推荐

最新文章

热门文章