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为美好的世界献上 bug

2834. Find the minimum sum of beautiful arrays.

2834. Find the Minimum Possible Sum of a Beautiful Array#

Given two positive integers: nn and targettarget.

If an array numsnums satisfies the following conditions, it is called a beautiful array.

  • nums.length==nnums.length == n
  • numsnums consists of positive integers that are pairwise distinct.
  • Within the range [0,n1][0, n-1], there does not exist two different indices ii and jj such that nums[i]+nums[j]==targetnums[i] + nums[j] == target.

Return the minimum sum that a beautiful array satisfying the conditions can have, and take the result modulo 109+710^9+7.

Example 1:

Input: n = 2, target = 3
Output: 4
Explanation: nums = [1,3] is a beautiful array.
- The length of nums is n = 2.
- nums consists of positive integers that are pairwise distinct.
- There does not exist two different indices i and j such that nums[i] + nums[j] == 3.
It can be proven that 4 is the minimum sum that a beautiful array satisfying the conditions can have.

Example 2:

Input: n = 3, target = 3
Output: 8
Explanation:
nums = [1,3,4] is a beautiful array.
- The length of nums is n = 3.
- nums consists of positive integers that are pairwise distinct.
- There does not exist two different indices i and j such that nums[i] + nums[j] == 3.
It can be proven that 8 is the minimum sum that a beautiful array satisfying the conditions can have.

Example 3:

Input: n = 1, target = 1
Output: 1
Explanation: nums = [1] is a beautiful array.

Constraints:

  • 1<=n<=1091 <= n <= 10^9
  • 1<=target<=1091 <= target <= 10^9

Greedy#

class Solution {
public:
    int minimumPossibleSum(int n, int target) {
        const int mod = 1e9 + 7;
        int m = target / 2;
        if (n <= m) {
            return (long long) (1 + n) * n / 2 % mod;
        }
        return ((long long) (1 + m) * m / 2 + 
                ((long long) target + target + (n - m) - 1) * (n - m) / 2) % mod;
    }
};
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