目录

• 题目截图
• 题目分析
• ac code
• 总结

题目截图

题目分析

• 开头必定是prime，结尾必定是not prime
• k = 1特判
• 找到所有可能的结尾点（最后一个不考虑）
• 结尾点i必须满足s[i]->not prime, s[i + 1]->prime
• 设结尾点集合为x
• 0 <= x1 < x2 < … < xm <= n - 1 选k - 1个点
• 不妨设为b1, b2, … b(k - 1), 满足
• 1.b1 + 1 >= minLength
• 2.bi - b(i - 1) >= minLength, i belongs to [1, k - 1]
• 3.n - 1 - b(k - 1) >= minLength
• 那么，这大概可以考虑dp
• 变量有前i个结束点，第j个实际结束段，minLength为定值不考虑
• 那么，dp[i][j]表示第i个结束点作为第j个结束段点的个数
• 答案为所有满足i作为n - 1 - x[i] >= minLength，的dp[i][-1]之和
• 转移方程：对于dp[i][j]而言，找到所有的i2使得x[i] - x[i2] >= minLength，那么dp[i][j] = sum(dp[i2][j - 1])
• i2的找法显然是一个简单二分，那么sum(dp[i2][j])显然就是一个前缀和
• 也就是说，在得到当前dp[i][j]的时候要维护一个第j列的前缀和，减少一层复杂度

ac code

提交的代码： 17 小时前
语言： python3添加备注class Solution:def beautifulPartitions(self, s: str, k: int, minLength: int) -> int:MOD = 10 ** 9 + 7n = len(s)primes = {'2', '3', '5', '7'}if s[0] not in primes or s[-1] in primes:return 0if k == 1:return int(s[0] in primes and s[-1] not in primes and len(s) >= minLength)# 可能的结束点j, 0 <= j < n - 1, j非质数 j + 1质数end_points = []for j, v in enumerate(s):if j < n - 1:if s[j] not in primes and s[j + 1] in primes:end_points.append(j)#print(end_points)x = end_pointsm = len(end_points)if m < k - 1:return 0# 0 <= x1 < x2 < ... < xm <= n - 1 选k - 1个点# 不妨设为b1, b2, ... b(k - 1), 满足# 1.b1 + 1 >= minLength# 2.bi - b(i - 1) >= minLength, i belongs to [1, k - 1]# 3.n - 1 - b(k - 1) >= minLength# dp[i][j]表示第i个结束点作为第j个结束段点的个数# dp[i][k]求和，其中n - 1 - x[i] >= minLengthdp = [[0] * (k - 1) for _ in range(m)]col_preSum = [[0] * (k - 1) for _ in range(m)] # 每列的前缀和优化for i in range(m): #initif x[i] + 1 >= minLength:#print(i, k, m)dp[i][0] = 1if i == 0:col_preSum[i][0] = dp[i][0]else:col_preSum[i][0] = col_preSum[i - 1][0] + dp[i][0]# 20221120我觉得大概能做出来把for i in range(m):for j in range(1, k - 1):# 本质上是一个j - 1列的前缀和# for i2 in range(i):#     if x[i] - x[i2] >= minLength:#         dp[i][j] += dp[i2][j - 1]#         dp[i][j] %= MODcheck_point = x[i] - minLengthi2 = bisect_right(x, check_point)if i2 == 0: # 找不到dp[i][j] == 0else: # 找到dp[i][j] += col_preSum[i2 - 1][j - 1]if i == 0:col_preSum[i][j] = dp[i][j]else:col_preSum[i][j] = col_preSum[i - 1][j] + dp[i][j]ans = 0for i in range(m):if n - 1 - x[i] >= minLength:ans += dp[i][-1]ans %= MODreturn ans % MOD

总结

• 预处理找分割点 + dp定义二维 + 前缀和优化
• 从这道题我看到了自己的进步。。。