目录
E - Lucky bag
题目大意:
思路解析:
代码实现:
E - Lucky bag
题目大意:
思路解析:
在方差中平均值只与输入有关为定值。看到数据范围为 2 <= D <= N <= 15,想到是否能使用状压dp来进行解答。
dp[i][j] (i为二进制)表示 i二进制状态下选择了这么多个物品,使用j个背包能够达到最小的方差。
代码实现:
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
static int mod1 = (int) 1e9 + 7;
static int mod2 = 998244353;
static int BD = 500;
public static void main(String[] args) throws IOException {
int n = input.nextInt();
int d = input.nextInt();
int[] arr = new int[n];
double sum = 0;
for (int i = 0; i < n; i++) {
arr[i] = input.nextInt();
sum += arr[i];
}
sum /= d;
double[][] dp = new double[(1 << n)][d + 1];
for (int i = 0; i < (1 << n); i++) {
double y = 0;
for (int j = 0; j < n; j++) {
if ((i & (1 <<j)) != 0)
y+=arr[j];
}
dp[i][1] = Math.pow(y-sum, 2);
for (int j = 2; j <= d; j++) {
dp[i][j] = dp[i][j-1] + dp[0][1];
int x = i;
while (x > 0){
dp[i][j] = Math.min(dp[i][j], dp[i-x][j-1] + dp[x][1]);
x = (x - 1) & i;
}
}
}
out.printf("%.15f",dp[(1 << n) - 1][d] / d);
out.flush();
out.close();
br.close();
}
// -------------------------------- 模板 ---------------------------
static boolean nextPermutation(int[] arr) { // 排列数循环模板 记得使用 do while 循环
int len = arr.length;
int left = len - 2;
while (left >= 0 && arr[left] >= arr[left + 1]) left--; // 从升序 一直往降序排列。
if (left < 0) return false;
int right = len - 1;
// 找到第一个升序的位置,将其改为降序。
while (arr[left] >= arr[right]) right--;
{
int t = arr[left];
arr[left] = arr[right];
arr[right] = t;
}
// 修改后它的前面仍然为降序,将前面全部修改为升序,这样能保证不会漏算,也不会算重
left++;
right = len - 1;
while (left < right) {
{
int t = arr[left];
arr[left] = arr[right];
arr[right] = t;
}
left++;
right--;
}
// System.out.println(Arrays.toString(arr));
return true;
}
public static long qkm(long a, long b, long mod) { // 快速幂模板
long res = 1;
while (b > 0) {
if ((b & 1) == 1) res = (res * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return res;
}
// // 线段树模板
// public static int lowbit(int x){
// return x & (-x);
// }
//
// public static void bulid(int n){
// for (int i = 1; i <= n; i++) {
// t[i] = a[i];
// int j = i + lowbit(i);
// if (j<=n) t[j] += t[i];
// }
// }
//
// public static void updata(int x, int val){
// while (x <= n){
// t[x] += val;
// x += lowbit(x);
// }
// }
//
// public static int query(int l, int r){
// int res = 0;
// while (l <= r){
// res += a[r];
// r--;
// while (r - lowbit(r) >= l){
// res += t[r];
// r -= lowbit(r);
// }
// }
// return res;
// }
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
static Input input = new Input(System.in);
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static class Input {
public BufferedReader reader;
public StringTokenizer tokenizer;
public Input(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public String nextLine() {
String str = null;
try {
str = reader.readLine();
} catch (IOException e) {
// TODO 自动生成的 catch 块
e.printStackTrace();
}
return str;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public Double nextDouble() {
return Double.parseDouble(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
}
}