CF 435B Little Pony and Harmony Chest

Little Pony and Harmony Chest

题解：

因为 1 <= ai <= 30 所以  1 <= bi <= 58， 因为 59 和 1 等效， 所以不需要59。

[1, 58]只有16个质数，对于这16个质数去状压。

对于1->58的数，我们计算出每个数对于质数来说的状态，然后转移。

代码：

#include<bits/stdc++.h>

using namespace std;

#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);

#define LL long long

#define ULL unsigned LL

#define fi first

#define se second

#define pb push_back

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define lch(x) tr[x].son[0]

#define rch(x) tr[x].son[1]

#define max3(a,b,c) max(a,max(b,c))

#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;

const int inf = 0x3f3f3f3f;

const int _inf = 0xc0c0c0c0;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const LL _INF = 0xc0c0c0c0c0c0c0c0;

const LL mod =  (int)1e9+;

const int N = 2e5 + ;

int prim[] = {,,,,,,,,,,,,,,,};

int state[];

int n;

int a[];

int dp[][<<];

int pre[][<<];

int use[][<<];

void P(int x, int v){

    if(!x) return;

    P(x-, pre[x][v]);

    printf("%d ", use[x][v]);

}

int Ac(){

    scanf("%d", &n);

    for(int i = ; i <= n; ++i) scanf("%d", &a[i]);

    int tot = <<;

    for(int i = ; i <= ; ++i)

        for(int j = ; j < ; ++j)

            if(i%prim[j] == )

                state[i] |=  << j;

    memset(dp, inf, sizeof dp);

    dp[][] = ;

    for(int i = ; i < n; ++i){

        for(int j = ; j < tot; ++j){

            if(dp[i][j] == inf) continue;

            for(int k = ; k <= ; ++k){

                if(j & state[k]) continue;

                int x = j | state[k];

                if(dp[i+][x] > dp[i][j] + abs(a[i+] - k)){

                    dp[i+][x] = dp[i][j] + abs(a[i+] - k);

                    pre[i+][x] = j;

                    use[i+][x] = k;

                }

            }

        }

    }

    int id = ;

    for(int j = ; j < tot; ++j){

        if(dp[n][id] > dp[n][j]){

            id = j;

        }

    }

    P(n, id);

    return ;

}

int main(){

    Ac();

    return ;

}