Codeforces 416D Population Size

Population Size

题意： 一共n个数， 每个-1都可以变成一个正数， 现在要求最少数目的等差子序列，并且在这个子序列必须要连着截取一段，不能分开截取。 样例1： 8 6 4 2 1 4 7 10 2 可以分成 { 8 6 4 2} {1 4 7 10 } {2} 3个等差子序列。

题解： 每一个序列都尽可能有更多的个数，贪心的去取就好了。

代码

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 const int N = 2e5+;
 ll n;
 ll a[N];
 int main()
 {
     ios::sync_with_stdio(false);
     cin.tie();
     cout.tie();
     while(cin >> n)
  {
     for(int i = ; i <= n; i++)
         cin >> a[i];
     a[n+] = -;
     bool star = , ok = ;
     ll cnt = , d = , ans = ;
     for(int i = ; i <= n; i++)
     {
         if(a[i] == - && !ok && star)
             cnt++;
         else if(a[i] == - && ok)
         {
             if(a[i-] + d > ) a[i] = a[i-] + d;//如果前面有等差序列
             else ans++, cnt = , star = , ok = ;//那么就将这个-1放进去
         }
         else if(a[i] == -  && !star)
         {
             int j = i+;
             while(a[j] == -) j++;
             if(a[j] == -)
                 break;
             ll dis = j+-i;
             ll diff = a[j] - a[i-];
             if(diff % dis == )// a[i-1]!=-1 a[j]!=-1 且有差等
             {　　　　　　　　　　//那么a[i]-a[j-1]内的-1都可以被代替成合法数
                 ok = ;
                 d = diff / dis;
                 i = j;
             }
             else ok = , star = , i = j-, ans++;
         }
         else if(a[i] != - && cnt)
         {
             int j = i+;
             while(a[j] == -) j++;
             if(a[j] == -) break;
             ll dis = j-i;
             ll diff = a[j]-a[i];
             if(diff % dis == )
             {
                 d = diff / dis;
                 if(a[i]-d*cnt > ) ok = , star = , i = j;
                 else i = j-, ok = , star = , ans++;
             }
             else i = j-, ok = , star = , ans++;
             cnt = ;
         }
         else if(star)
             star = ;
         else if(!ok)
             d = a[i] - a[i-], ok = ;
         else if(a[i-]+d != a[i])
             ans++, star = , ok = , i--;
     }
     cout << ans +  << endl;//最后答案要加一，因为最后至少有一个序列
 }
     return ;
 }