CF-416D-（贪心+思维）

题意：https://codeforces.com/problemset/problem/416/D

合成连续等差数列，-1可替换任何数字，问最少多少段

思路：

一开始dp，d得头都痛了。

题解说每次取两个正数搞搞，差不多：能合体就合体，然后往后拖（还说反正不会使答案更差。。。

具体见代码

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr strcat
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
#include <iomanip>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
clock_t __START,__END;
double __TOTALTIME;
void _MS(){__START=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
const db E=2.718281828;
const db PI=acos(-1.0);
const ll INF=(1LL<<60);
const int inf=(1<<30);
const db ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;
#define int ll

int a[N];

signed main()
{
    int n,ans=0;
    sc("%lld",&n);
    for(int i=1;i<=n;++i)sc("%lld",&a[i]);
    int pos=1;
    int i,j;
    while(pos<=n)
    {
        ans++;
        for(i=pos;a[i]==-1;++i);
        for(j=i+1;a[j]==-1;++j);
        if(j>n)break;
        int d=(a[j]-a[i])/(j-i);
        if((a[j]-a[i])%(j-i)||a[i]-(i-pos)*d<=0)//不能和前面一个数合体,前一个数想拖多长拖多长
        {
            pos=j;
            continue;
        }
        pos=j+1;
        while(pos<=n&&a[j]+d*(pos-j)>0&&(a[pos]==-1||a[pos]==a[j]+d*(pos-j)))//能往后拖多长就拖多长
            pos++;
    }
    pr("%lld
",ans);
    return 0;
}

/**************************************************************************************/