Codeforces 416D Population Size

Posted 2020-10-24 Schenker

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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N = 2e5+5;
 5 ll n;
 6 ll a[N];
 7 int main()
 8 {
 9     ios::sync_with_stdio(false);
10     cin.tie(0);
11     cout.tie(0);
12     while(cin >> n)
13  {
14     for(int i = 1; i <= n; i++)
15         cin >> a[i];
16     a[n+1] = -2;
17     bool star = 1, ok = 0;
18     ll cnt = 0, d = 0, ans = 0;
19     for(int i = 1; i <= n; i++)
20     {
21         if(a[i] == -1 && !ok && star)
22             cnt++;
23         else if(a[i] == -1 && ok)
24         {
25             if(a[i-1] + d > 0) a[i] = a[i-1] + d;//如果前面有等差序列
26             else ans++, cnt = 1, star = 1, ok = 0;//那么就将这个-1放进去
27         }
28         else if(a[i] == -1  && !star)
29         {
30             int j = i+1;
31             while(a[j] == -1) j++;
32             if(a[j] == -2)
33                 break;
34             ll dis = j+1-i;
35             ll diff = a[j] - a[i-1];
36             if(diff % dis == 0)// a[i-1]!=-1 a[j]!=-1 且有差等
37             {　　　　　　　　　　//那么a[i]-a[j-1]内的-1都可以被代替成合法数
38                 ok = 1;
39                 d = diff / dis;
40                 i = j;
41             }
42             else ok = 0, star = 1, i = j-1, ans++;
43         }
44         else if(a[i] != -1 && cnt)
45         {
46             int j = i+1;
47             while(a[j] == -1) j++;
48             if(a[j] == -2) break;
49             ll dis = j-i;
50             ll diff = a[j]-a[i];
51             if(diff % dis == 0)
52             {
53                 d = diff / dis;
54                 if(a[i]-d*cnt > 0) ok = 1, star = 0, i = j;
55                 else i = j-1, ok = 0, star = 1, ans++;
56             }
57             else i = j-1, ok = 0, star = 1, ans++;
58             cnt = 0;
59         }
60         else if(star)
61             star = 0;
62         else if(!ok)
63             d = a[i] - a[i-1], ok = 1;
64         else if(a[i-1]+d != a[i])
65             ans++, star = 1, ok = 0, i--;
66     }
67     cout << ans + 1 << endl;//最后答案要加一，因为最后至少有一个序列
68 }
69     return 0;
70 }