Codeforces 453B Little Pony and Harmony Chest(状压)

题目链接：Codeforces 453B Little Pony and Harmony Chest

题目大意：给定一个序列a， 求一序列b，要求∑|ai?bi|最小。并且b中任意两数的最大公约束为1.

解题思路：因为b中不可能含有相同的因子，所以每个素数只能使用1次。又因为说ai最大为30，所以素数只需要考虑到57即可。因为即使对于30而言，59和1的代价是一样的。

所以有dp[i][j]表示的是到第i个数，使用过的素数j。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int maxn = 105;
const int sign[maxn] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};
const int INF = 0x3f3f3f3f;
const int sn = 1<<16;

int N, num[maxn], s[maxn];
int dp[maxn][sn+5], rec[maxn][sn+5];

inline int getstate (int a) {
    int u = 0;
    for (int i = 0; i < 16; i++) {
        while (a % sign[i] == 0) {
            a /= sign[i];
            u |= (1<<i);
        }
    }
    return u;
}

inline void put_ans (int d, int S) {
    if (d == 0)
        return;

    int u = rec[d][S];
    put_ans(d-1, S^s[u]);
    num[d] = u;
}

void solve () {
    memset(rec, -1, sizeof(rec));
    rec[0][0] = 0;

    for (int i = 0; i < N; i++) {
        for (int j = 0; j < sn; j++) {
            if (rec[i][j] == -1)
                continue;

            for (int k = 1; k < 60; k++) {
                if (j&s[k])
                    continue;

                int v = j|s[k];
                if (rec[i+1][v] == -1 || dp[i][j] + abs(k-num[i+1]) < dp[i+1][v]) {
                    rec[i+1][v] = k;
                    dp[i+1][v] = dp[i][j] + abs(k-num[i+1]);
                }
            }
        }
    }

    int ans = INF, id;
    for (int i = 0; i < sn; i++) {
        if (dp[N][i] < ans && rec[N][i] != -1) {
            ans = dp[N][i];
            id = i;
        }
    }

    put_ans(N, id);
    for (int i = 1; i < N; i++)
        printf("%d ", num[i]);
    printf("%d\n", num[N]);

}

int main () {
    for (int i = 1; i < 60; i++)
        s[i] = getstate(i);

    scanf("%d", &N);
    for (int i = 1; i <= N; i++)
        scanf("%d", &num[i]);

    solve();
    return 0;
}

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