uestc 稳住GCD

DP

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<string>
 6 #include<queue>
 7 #include<algorithm>
 8 #include<map>
 9 #include<iomanip>
10 #include<climits>
11 #include<string.h>
12 #include<cmath>
13 #include<stdlib.h>
14 #include<vector>
15 #define INF 1e7
16 #define MAXN 111111
17 #define Mod 1000007
18 using namespace std;
19 typedef long long LL;
20
21 int dp[777][10010];
22 int a[777];
23 int n, g, i, j;
24
25 int gcd(int a, int b)
26 {
27     return b == 0 ? a : gcd(b,a%b);
28 }
29
30 void init()
31 {
32     g = 0;
33     for (i = 1; i <= n; ++i)
34         for (j = 0; j <= 10004; ++j)
35             dp[i][j] = Mod;
36 }
37 int main()
38 {
39     cin >> n;
40     init();
41     int maxi = 1;
42     for (i = 1; i <= n; ++i) {
43         cin >> a[i];
44         g = gcd(a[i],g);
45         maxi = max(maxi, a[i]);
46         dp[i][a[i]] = 1;
47     }
48     for (i = 2; i <= n; ++i) {
49         for (j = 0; j <= maxi; ++j) {
50             if (dp[i - 1][j] != Mod) {
51                 int gg = gcd(a[i],j);
52                 if (gg == j)
53                     dp[i][gg] = min(dp[i][gg], dp[i - 1][j]);
54                 else
55                     dp[i][gg] = min(dp[i][gg], dp[i - 1][j] + 1);
56             }
57         }
58     }
59     cout << n - dp[n][g] << endl;
60     return 0;
61 }