Codeforces 453B Little Pony and Harmony Chest(状压)

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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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