


codeforces Round #259(div2) DProblem Solution Report_html/css_WEB-ITnose
Jun 24, 2016 am 11:55 AM
D. Little Pony and Harmony Chest
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Princess Twilight went to Celestia and Luna's old castle to research the chest from the Elements of Harmony.
A sequence of positive integers?bi?is harmony if and only if for every two elements of the sequence their greatest common divisor equals 1. According to an ancient book, the key of the chest is a harmony sequence?bi?which minimizes the following expression:
You are given sequence?ai, help Princess Twilight to find the key.
Input
The first line contains an integer?n?(1?≤?n?≤?100) ? the number of elements of the sequences?a?and?b. The next line contains?n?integersa1,?a2,?...,?an?(1?≤?ai?≤?30).
Output
Output the key ? sequence?bi?that minimizes the sum described above. If there are multiple optimal sequences, you can output any of them.
Sample test(s)
input
51 1 1 1 1
output
1 1 1 1 1
input
51 6 4 2 8
output
1 5 3 1 8
題目大意:
給出N個數(shù)ai,求出另一個序列bi,要求sum |ai-bi|,最短,且所有的bi都互質(zhì)。
解法:
這里題目給了幾個很顯眼的條件,ai限制在了1~30之間,由于可以bi無限選1這個數(shù),那么|ai-bi| 最大就是29了,意味著bi < 59的。
要求所有的bi互質(zhì),可以化為所有的bi分解出來的質(zhì)因數(shù)均不相同,bi < 59,有16個質(zhì)數(shù)。這里我們很容易聯(lián)想到狀態(tài)壓縮DP了。
用s表示當(dāng)前階段用了哪些質(zhì)因數(shù)的狀態(tài),例如 s = 3 = 11 代表目前狀態(tài)下使用了第一個和第二個質(zhì)因數(shù)。
很快我們就可以寫出狀態(tài)轉(zhuǎn)移方程:
? ? ? ?f[i][s] = min(f[i-1][s^c[k]] abs(a[i] - k))。 ? ?其中c[k]表示數(shù)字k使用了哪些質(zhì)因數(shù)。
代碼:
#include#include #include #define M_max 60#define N_max 123#define inf 0x3f3f3f3fusing namespace std;int p[N_max], c[M_max], a[N_max];int f[N_max][1<<16], pre[N_max][1<<16][2];int n, cnt, minnum, minpos;void prime() { for (int i = 2; i <= M_max; i++) { bool flag = false; for (int j = 2; j <= sqrt(i); j++) if (i%j == 0) { flag = true; break; } if (!flag) p[++cnt] = i; } for (int i = 1; i <= M_max; i++) for (int j = 1; j <= cnt; j++) if (i%p[j] == 0) c[i] |= 1 << (j-1);}void init() { prime(); scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", &a[i]);}void print(int x, int pos) { if (x == 0) return; print(x-1, pre[x][pos][0]); printf("%d ", pre[x][pos][1]);}void solve() { memset(f, inf, sizeof(f)); memset(f[0], 0, sizeof(f[0])); minnum = inf; for (int i = 1; i <= n; i++) for (int s = 0; s < (1<<16); s++) for (int k = 1; k <= M_max; k++) if ((s&c[k]) == c[k]) { int tmp = f[i-1][s^c[k]] + abs(a[i]-k); if (tmp < f[i][s]) { f[i][s] = tmp; pre[i][s][0] = s^c[k]; pre[i][s][1] = k; } } for (int s = 0; s < (1<<16); s++) if (f[n][s] < minnum) { minnum = f[n][s]; minpos = s; } print(n, minpos);}int main() { init(); solve();}

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