题解 | 单组_spj判断数组之和
单组_spj判断数组之和
https://www.nowcoder.com/practice/da783089da3b4a3188d240e1b7ac4b23
n, m = map(int, input().split())
# 前n-1个元素为1,最后一个元素为1 + (m - n)
arr = [1] * (n - 1) + [m - n + 1]
print(' '.join(map(str, arr)))
单组_spj判断数组之和
https://www.nowcoder.com/practice/da783089da3b4a3188d240e1b7ac4b23
n, m = map(int, input().split())
# 前n-1个元素为1,最后一个元素为1 + (m - n)
arr = [1] * (n - 1) + [m - n + 1]
print(' '.join(map(str, arr)))
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