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#线性dp，排列组合#洛谷 2476 [SCOI2008]着色方案
题目 分析（弱化版） 最暴力的想法就是直接维护每种颜色的个数dp， 弱化版有一个很突出的地方就是 \(c_i\leq 5\)， 也就是说可以将相同个数的颜色合并按照个数dp， 设 \(dp[c1][c2][c3][c4][c5][las]\) 表示个数为 \(i\) 的颜色有 \(ci\) 种，并且上一次选了个数为 \(las\) 的颜色的方案数String painter HDU - 2476
原题链接 考察:区间dp 这道题可以想到P4170涂色和Acwing 编辑距离的结合. 错误思路: 如果直接按a[i]==b[i]与a[j]==b[j]划分的话(实际上a[i]==b[i]与j可以合并),就会少了考虑b连续的条件.使得次数增加了. 错误思路2: 将a[i]==a[j]&&b[i]==b[j]与b[i]==a[i]合并LOJ#2476. 「2018 集训队互测 Day 3」蒜头的奖杯
题面 题解 设 \(\mathbf f' = \mathbf f * \mu\)，\(G_{\mathbf f} (n) = \sum_{n | d} \mathbf f(d)\)，记 \((i, j) = \gcd(i, j)\)，\([i, j] = \operatorname{lcm}(i, j)\)。 令 \(S_i=A_{id}, T_i = B_{id}, W_i = D_{id}, m = \left\lfloor \frac nd \riHDU-2476 String painter
There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other character you want. Thhdu 2476 String painter 区间DP
String painter Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6988 Accepted Submission(s): 3381 Problem Description There are two strings A and B with equal length. Both strings are made up of