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======codeforces round 658 div2======
=====A Common Subsequence======
题意：给两个串，找出最小的非空相同字串。
题解：最小非空字串即1个的情况，否则一定找不到，找在两个串都出现过的数即可。
=====B Sequential Nim=====
题意：两个人做博弈，给一串数，从第一个数字开始，每个人轮流拿走一个正数，拿走后剩余数字会减少那么多，当出现0，之后开始拿下一个，无法经行操作的人获胜。
题解：若一串数字都不是1，则先手必胜，因为先手每次可以将一个数字拿的只剩下1，然后打最后一个数字时将最后一个数全部拿走，所以一定会赢。静儿可以推出，一个拿到非1数字的人获胜，因为它可以控制赛局而让自己最后获胜。而出现1，则相当于交换先手顺序，所以只要看谁第一个拿到非1数即可。