A. Long Comparison
题解
首先我们可以比较x的位数，如果位数不同，我们就对位数较少的数不断乘以10，直到位数相通。然后再比较调整后的p的大小。如果p一样，那么再比较x。
代码
Java
Submission #138269313 - Codeforces
Codeforces. Programming competitions and contests, programming community
C++
Submission #138269423 - Codeforces
Codeforces. Programming competitions and contests, programming community
B. Absent Remainder
题解
显然x mod y < x。因此我们将x永远选定为数组的最小值，然后y随便从剩下的数中选即可。
代码
Java
Submission #138271750 - Codeforces
Codeforces. Programming competitions and contests, programming community
C++
Submission #138271868 - Codeforces
Codeforces. Programming competitions and contests, programming community
C. Poisoned Dagger
题解
二分搜索k的值。每次按照题目描述计算总伤害即可。
代码
Java
Submission #138273665 - Codeforces
Codeforces. Programming competitions and contests, programming community
C++
Submission #138274417 - Codeforces
Codeforces. Programming competitions and contests, programming community
D. MEX Sequences
这个题目需要想清楚有那些构造方式。有了构造的方法dp就会显得比较显然。
Educational Codeforces Round 118 MEX Sequences 题解 (Java/C++)
题解首先我们考虑有那些选择方式可以满足条件。显然[0, 1, 2, 3, 4, 5]是符合条件的，因为对于任意的i，x[i]-MEX(x[1],..., x[i])=-1。同理，[0, 1, 2, 3, 4, 5, 5, ... ,5]也是满足条件的。
E. Crazy Robot
题解
从L出发，dfs即可。每次寻找可能为+号的点，然后继续搜索即可。
而这个题目C++反而比Java容易TLE。因为n和m的值变动范围较大，而C++动态生成二维数组并初始化的速度似乎没有Java快。
代码
Java
Submission #138360324 - Codeforces
Codeforces. Programming competitions and contests, programming community
C++
Submission #138364073 - Codeforces
Codeforces. Programming competitions and contests, programming community