Educational Codeforces Round 118 (Rated for Div. 2)
solved: 3/6
A
把x1 x2補到相同的位數之後再比較p1p2的大小, 如果p1=p2, 則比較x1, x2的大小
實際上就是科學計數法
B
若a mod b=c 則 $c \in [0, b)$
所以排序之後對最小值取模肯定能得到序列中沒有的數
C
二分答案, 迴圈模擬一下判斷就行
如果持續時間小的能幹掉龍, 那大的肯定也可以
補題1/3
D
觀察條件, 每加入一個數, 當前的mex = x-1 x+1,
mex代表子序列中有1..mex-1中的所有數
再考慮到如果k>mex+1 那麼他不可能出現在序列中, 而mex肯定是遞增的,
所以可能的序列就只有三種情況
不降的: 0000(若干個)111(若干個)...
不降接抖動的: 000(若干個)....(n+2)(若干個)n(若干個)n+2(若干個)
注意特判1的情況, 全為1的情況是合法的
總之就是個分類討論題, 分類到最後只能得到一個不降的結論, 賽後交了好多發才過T_T
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