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  • 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 }
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  • 原文地址:https://www.cnblogs.com/usedrosee/p/4287034.html
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