1354. Construct Target Array With Multiple Sums

Given an array of integers target. From a starting array, A consisting of all 1's, you may perform the following procedure :

• let x be the sum of all elements currently in your array.

• choose index i, such that 0 <= i < target.size and set the value of A at index i to x.

• You may repeat this procedure as many times as needed.

Return True if it is possible to construct the target array from A otherwise return False.

Example 1:

Input: target = [9,3,5]
Output: true
Explanation: Start with [1, 1, 1] 
[1, 1, 1], sum = 3 choose index 1
[1, 3, 1], sum = 5 choose index 2
[1, 3, 5], sum = 9 choose index 0
[9, 3, 5] Done

Example 2:

Input: target = [1,1,1,2]
Output: false
Explanation: Impossible to create target array from [1,1,1,1].

Example 3:

Input: target = [8,5]
Output: true

Constraints:

• N == target.length

• 1 <= target.length <= 5 * 10^4

• 1 <= target[i] <= 10^9

从全1数组开始，每步计算数组的和并用该和将一个元素值替换，问最后能否变成给定的target数组。此题和780 Reach Points极为类似，倒着从target数组开始，被替换的元素一定是当前最大的元素并且它的值是上一步的数组和，再根据当前和就可以算出该元素替换前的值是多少，依次类推看最后能否退到1。对于像[100000000,1]的，最大值是其它元素很多倍的，取mod来进一步压缩时间。

参考自这。

class Solution {
public:
    bool isPossible(vector<int>& target) {
        long s = accumulate(target.begin(), target.end(), (long) 0);
        priority_queue<int> pq(target.begin(), target.end());
        while (true) {
            auto cur = pq.top(); pq.pop();
            if (cur == 1) return true;
            auto rest = s - cur;
            if (cur <= rest) return false;
            auto pre = cur % rest;
            pq.push(pre);
            s = pre + rest;
        }
    }
};