1590.-make-sum-divisible-by-p.md

1590. Make Sum Divisible by P

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class Solution:
    def minSubarray(self, A: List[int], p: int) -> int:
        
        ## 等价于　
        ## 得到全数组的余数是多少
        need = sum(A)%p
        
        ##dp　代表　余数和ｉｄｘ
        last = {0: -1}
        res = n = len(A)
        cur = 0
        for idx, v in enumerate(A):
            cur = (cur + v)%p
            last[cur] = idx
            ## python mod　会变正数
            if (cur - need + p)%p in last:
                res = min(res, idx - last[(cur - need + p)%p])
                
        return -1 if res == n else res

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