给定一个数组nums，将元素分为若干个组，使得每组和相等，求出满足条件的所有分组中，组内元素和的最小值。
第一行输入 m
接着输入m个数，表示此数组nums
数据范围：1 <= m <= 50, 1 <= nums[i] <= 50
最小拆分数组和
7
4 3 2 3 5 2 1
5
可以等分的情况有：
4 个子集（5），（1,4），（2,3），（2,3）
2 个子集（5, 1, 4），（2,3, 2,3）
但最小的为5。
这道题目要求将数组拆分为多个组，使每个组的和相等，并求出这样的组中最小的组和。以下是解题思路和复杂度分析：
时间复杂度：该算法的时间复杂度较高，最坏情况下为 O(n!)，其中 n 是数组长度。这是因为回溯法在穷举所有可能的分组组合时，时间复杂度较高。然而，实际问题中的数据规模较小（数组长度最大为 50），因此算法在实际问题中的运行时间仍然是可接受的。
空间复杂度：算法的空间复杂度主要取决于递归栈的深度，最坏情况下为 O(n)，因此空间复杂度为 O(n)。
这道题目要求找到一个数组中满足一定条件的分组，并求出这些分组中最小的组和。采用回溯法的思路，通过递归地搜索满足条件的分组组合，可以找到答案。虽然时间复杂度较高，但对于实际问题中的数据规模，算法的运行时间仍然是可接受的。
#include <stdio.h>
#include <stdlib.h>
int min_sum = 50 * 50 + 1;
int target;
int compare(const void *a, const void *b) {
return *(int *)b - *(int *)a;
}
void backtrack(int *nums, int index, int m, int sum, int num_groups, int total_sum) {
if (sum == target) {
total_sum += sum;
if (total_sum == target * num_groups) {
if (sum < min_sum) {
min_sum = sum;
}
}
else {
backtrack(nums, 0, m, 0, num_groups - 1, total_sum);
}
}
else {
for (int i = index; i < m; ++i) {
if (sum + nums[i] <= target) {
int temp = nums[i];
nums[i] = -1;
backtrack(nums, i + 1, m, sum + temp, num_groups, total_sum);
nums[i] = temp;
}
}
}
}
int main() {
int m;
scanf("%d", &m);
int nums[50];
int total = 0;
for (int i = 0; i < m; ++i) {
scanf("%d", &nums[i]);
total += nums[i];
}
qsort(nums, m, sizeof(int), compare);
for (int num_groups = 1; num_groups <= m; ++num_groups) {
if (total % num_groups == 0) {
target = total / num_groups;
backtrack(nums, 0, m, 0, num_groups, 0);
}
}
printf("%d\n", min_sum);
return 0;
}
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int min_sum = 50 * 50 + 1;
int target;
void backtrack(vector<int> &nums, int index, int m, int sum, int num_groups, int total_sum) {
if (sum == target) {
total_sum += sum;
if (total_sum == target * num_groups) {
if (sum < min_sum) {
min_sum = sum;
}
}
else {
backtrack(nums, 0, m, 0, num_groups - 1, total_sum);
}
}
else {
for (int i = index; i < m; ++i) {
if (sum + nums[i] <= target) {
int temp = nums[i];
nums[i] = -1;
backtrack(nums, i + 1, m, sum + temp, num_groups, total_sum);
nums[i] = temp;
}
}
}
}
int main() {
int m;
cin >> m;
vector<int> nums(m);
int total = 0;
for (int i = 0; i < m; ++i) {
cin >> nums[i];
total += nums[i];
}
sort(nums.rbegin(), nums.rend());
for (int num_groups = 1; num_groups <= m; ++num_groups) {
if (total % num_groups == 0) {
target = total / num_groups;
backtrack(nums, 0, m, 0, num_groups, 0);
}
}
cout << min_sum << endl;
return 0;
}
import java.util.Arrays;
import java.util.Scanner;
public class Main {
private static int minSum = 50 * 50 + 1;
private static int target;
private static void backtrack(int[] nums, int index, int m, int sum, int numGroups, int totalSum) {
if (sum == target) {
totalSum += sum;
if (totalSum == target * numGroups) {
if (sum < minSum) {
minSum = sum;
}
} else {
backtrack(nums, 0, m, 0, numGroups - 1, totalSum);
}
} else {
for (int i = index; i < m; ++i) {
if (sum + nums[i] <= target) {
int temp = nums[i];
nums[i] = -1;
backtrack(nums, i + 1, m, sum + temp, numGroups, totalSum);
nums[i] = temp;
}
}
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int m = scanner.nextInt();
int[] nums = new int[m];
int total = 0;
for (int i = 0; i < m; ++i) {
nums[i] = scanner.nextInt();
total += nums[i];
}
scanner.close();
Arrays.sort(nums);
for (int i = 0; i < m / 2; i++) {
int temp = nums[i];
nums[i] = nums[m - i - 1];
nums[m - i - 1] = temp;
}
for (int numGroups = 1; numGroups <= m; ++numGroups) {
if (total % numGroups == 0) {
target = total / numGroups;
backtrack(nums, 0, m, 0, numGroups, 0);
}
}
System.out.println(minSum);
}
}
from itertools import permutations
def min_group_sum(nums):
total = sum(nums)
min_sum = 50 * 50 + 1
def backtrack(index, m, sum, num_groups, total_sum):
nonlocal min_sum
if sum == target:
total_sum += sum
if total_sum == target * num_groups:
if sum < min_sum:
min_sum = sum
else:
backtrack(0, m, 0, num_groups - 1, total_sum)
else:
for i in range(index, m):
if sum + nums[i] <= target:
temp = nums[i]
nums[i] = -1
backtrack(i + 1, m, sum + temp, num_groups, total_sum)
nums[i] = temp
nums.sort(reverse=True)
m = len(nums)
for num_groups in range(1, m + 1):
if total % num_groups == 0:
target = total // num_groups
backtrack(0, m, 0, num_groups, 0)
return min_sum
if __name__ == "__main__":
nums = [4, 3, 2, 3, 5, 2, 1]
print(min_group_sum(nums))
const minGroupSum = (nums) => {
const total = nums.reduce((acc, val) => acc + val, 0);
let minSum = 50 * 50 + 1;
const backtrack = (index, m, sum, numGroups, totalSum) => {
if (sum === target) {
totalSum += sum;
if (totalSum === target * numGroups) {
if (sum < minSum) {
minSum = sum;
}
} else {
backtrack(0, m, 0, numGroups - 1, totalSum);
}
} else {
for (let i = index; i < m; ++i) {
if (sum + nums[i] <= target) {
const temp = nums[i];
nums[i] = -1;
backtrack(i + 1, m, sum + temp, numGroups, totalSum);
nums[i] = temp;
}
}
}
};
nums.sort((a, b) => b - a);
const m = nums.length;
for (let numGroups = 1; numGroups <= m; ++numGroups) {
if (total % numGroups === 0) {
target = total / numGroups;
backtrack(0, m, 0, numGroups, 0);
}
}
return minSum;
};
const nums = [4, 3, 2, 3, 5, 2, 1];
console.log(minGroupSum(nums));
const minGroupSum = (nums) => {
const total = nums.reduce((acc, val) => acc + val, 0);
let minSum = 50 * 50 + 1;
const backtrack = (index, m, sum, numGroups, totalSum) => {
if (sum === target) {
totalSum += sum;
if (totalSum === target * numGroups) {
if (sum < minSum) {
minSum = sum;
}
} else {
backtrack(0, m, 0, numGroups - 1, totalSum);
}
} else {
for (let i = index; i < m; ++i) {
if (sum + nums[i] <= target) {
const temp = nums[i];
nums[i] = -1;
backtrack(i + 1, m, sum + temp, numGroups, totalSum);
nums[i] = temp;
}
}
}
};
nums.sort((a, b) => b - a);
const m = nums.length;
for (let numGroups = 1; numGroups <= m; ++numGroups) {
if (total % numGroups === 0) {
target = total / numGroups;
backtrack(0, m, 0, numGroups, 0);
}
}
return minSum;
};
const nums = [4, 3, 2, 3, 5, 2, 1];
console.log(minGroupSum(nums));