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
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const db E=2.718281828; 45 const db PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const db ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=(int)1e6+10; 51 #define int ll 52 53 int a[N]; 54 55 signed main() 56 { 57 int n,ans=0; 58 sc("%lld",&n); 59 for(int i=1;i<=n;++i)sc("%lld",&a[i]); 60 int pos=1; 61 int i,j; 62 while(pos<=n) 63 { 64 ans++; 65 for(i=pos;a[i]==-1;++i); 66 for(j=i+1;a[j]==-1;++j); 67 if(j>n)break; 68 int d=(a[j]-a[i])/(j-i); 69 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 70 { 71 pos=j; 72 continue; 73 } 74 pos=j+1; 75 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 76 pos++; 77 } 78 pr("%lld ",ans); 79 return 0; 80 } 81 82 /**************************************************************************************/