题解 | #矩阵乘法计算量估算# 先找最左侧的')', 再找它左侧最近的'('。处理后的元素变成特殊标记。

vscode反复debug后终于通了。
debug时希望在窗口紧挨着的变量，最好头几个字母一样。
用栈做好像很简单，后面再补充

def f(w_hs,st):
    num_mul=0
    chrs=list(st)
    cids=[str(i) for i in range(len(st))]
    kuo_l=[]
    kuo_r=[]
    for id_c in range(len(st)):
        if st[id_c]=='(':
            kuo_l.append(id_c)
            w_hs.insert(id_c, 1)
        if st[id_c]==')':
            kuo_r .append(id_c)
            w_hs.insert(id_c, 1)
    n_kuo= len(kuo_l)
    #随便一个符号，做标记  用@比None好，debug时容易对齐
    while chrs.count(None) <len(chrs) -1 :
        for id_p in range(n_kuo):
            p_right= kuo_r[id_p] - 1
            chrs[kuo_r[id_p]] = None  #随便一个符号，做标记
            p_left= p_right - chrs[:p_right][::-1].index('(') -1
            chrs[p_left ] = None
            # id_c+1会取到right
            for id_c in range(p_left,p_right):
                if chrs[id_c] != None:
                    for id_c2 in range(id_c+1,p_right+1):
                        if chrs[id_c2] != None: 
                            num_mul += (w_hs[id_c][0]*w_hs[id_c][1]) * w_hs[id_c2][1]
                            # print('num_mul')
                            # print(num_mul)
                            chrs[id_c] = None
                            w_hs[id_c2] = (w_hs[id_c][0] , w_hs[id_c2][1] )
                            w_hs[id_c] = None
                            # 乘一次就break，起点右挪
                            break
    print(num_mul)

while 1:
    try:
        n=int(input())
        w_hs=[]
        for i_n in range(n):
            w_hs.append(  tuple(
                              map(
                                  int, input().strip().split()
                                  )
                             )
                      )

        st=input().strip()
        f(w_hs, st)
    except Exception as e:
        if not isinstance(e,EOFError):
            print('!!!!!!!!!!leo_you_have_Error!!!!!!!')
        break