dp=[0 for i in range(n)] #创建n*1的列表 dp[0]=1;dp[1]=2 #初始值 for i in range(2,n): dp[i]=dp[i-1]+dp[i-2]#递归式:状态转移方程 return dp[n-1] >>> g11(10) 89 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 二、二维DP ①求方案数 问题2.1 描述:在m*n的方格...
n,m=map(int,input().split()) v=[0 for i in range(0,n+1)] w=[0 for i in range(0,n+1)] dp=[[0 for i in range(0,m+1)] for i in range(0,n+1)] for i in range(0,n): v[i],w[i]=map(int,input().split()) for i in range(0,n): for j in range(0,m+1...
def uniquePaths(self, m, n): dp=[] for i in range(n):#初始化 dp.append([0]*m) for i in range(n): dp[i][0]=1 dp[0]=[1]*m for i in range(1,n): for j in range(1,m): dp[i][j]=dp[i-1][j]+dp[i][j-1] return dp[n-1][m-1]...
dp = [[0] * (Vbag + 1) for _ in range(num+1)] ap = [[0] * (Vbag + 1)] * (num + 1) print(ap == dp) # weight = n # value = weight weight = [0, 2, 3, 4, 5] value = [0, 3, 4, 5, 6] for i in range(1, num + 1): for j in range(1, Vbag + 1...
dp = [[0] * n] * m for 循环 dp= [[0for_inrange(n)]for_inrange(m )] 乘法+for循环 dp = [[0] * nfor_inrange(m )] 但是,在使用列表乘法创建的时候会出现问题: dp = [[0]*2]*2print(dp)# [[0, 0], [0, 0]]print(id(dp[0]))#3142462401856print(id(dp[1]))#31424624018...
dp[0]=0foriinrange(mx):# 如果 dp[i] 为 -1 ,则使用状态 i 中的火柴棍,# 无法拼出满足题意的正方形的边,直接处理下一个状态ifdp[i]==-1:continue# 枚举接下来要使用的火柴棍forjinrange(n):# 如果第 j 根火柴棍不在状态 i 中,且加入后拼出的边长不超过 target ,# 则状态 i | (1 << ...
# #钢条切割问题:自低向上(由小到大) # #获得最大值 def max(a,b): maxData = a; if maxData < b: maxData = b; return maxData def BOTTOM_UP_CUT_ROD(p,n): r = {} r[0] = 0 for j in range(1,n): q = 0 for i in range(1,j): q = max(int(q),int(p[i] + r[...
foriinrange(n - 2): num = f[-1] + f[-2] f.append(num) returnf[n] print(fibnacci_n_recurision(10)) 实际上,带「备忘录」的递归算法,把一棵存在巨量冗余的递归树通过「剪枝」,改造成了一幅不存在冗余的递归图,极大减少了子问题(即递归图中节点)的个数。
zeros((n, n)) for i in range(n): for j in range(n): if i != j: sim[i][j] = np.corrcoef(ratings[i], ratings[j])[0, 1] return sim # 基于用户的协同过滤推荐算法 def user_based_collaborative_filtering(ratings, threshold=0.6): n = ratings.shape[0] sim = similarity(ratings)...