字节跳动笔试第3题

def main(l,r):     sub = list(seq[l:r + 1])     i = 0     score = 0     stack = []     flag = ''     while 0 <= i < len(sub):         ####### 还未有标志位的处理         if flag == "" and sub[i] not in "<>":             score += int(sub[i])             sub[i] = str(int(sub[i]) - 1)             stack.append(int(sub[i]))             i += 1         elif flag == "" and sub[i] == "<":             score += sum(stack)             return score         elif flag == '' and sub[i] == ">":             flag += sub[i]             idx = i             i += 1             stack = []         ########## 标志位为">"         elif flag == ">" and sub[i] not in "<>":             stack.append(int(sub[i]))             sub[i] = str(int(sub[i]) - 1)             i += 1             if i == len(sub):                 score += sum(stack)                 return score         elif flag == ">" and sub[i] == ">":             score += sum(stack)             stack = []             idx = i             i += 1         elif flag == ">" and sub[i] == "<":             # 根据stack里最大的数值来判断删除">"还是"<"             maxx = max(stack)             for j in stack:                 tmp = ((1 + j) * j) // 2                 score += tmp             if maxx % 2 == 0:                 sub = sub[:idx + 1] + sub[i + 1:]                 i = idx + 1                 flag = ">"             else:                 sub = sub[:idx] + sub[i:]                 i = idx - 1                 flag = "<"             stack = []         ########### 标志位为"<"         elif flag == "<" and sub[i] not in "<>":             stack.append(int(sub[i]))             sub[i] = str(int(sub[i]) - 1)             i -= 1             if i == -1:                 score += sum(stack)                 return score         elif flag == "<" and sub[i] == "<":             score += sum(stack)             stack = []             i -= 1         elif flag == "<" and sub[i] == ">":             # 根据stack里最大的数值来判断删除">"还是"<"             maxx = max(stack)             for j in stack:                 tmp = ((1 + j) * j) // 2                 score += tmp             if maxx % 2 != 0:                 sub = sub[:i + 1] + sub[idx + 1:]                 i += 1                 flag = ">"             else:                 sub = sub[:i] + sub[idx:]                 i -= 1                 flag = "<"             stack = []     return score if __name__ == '__main__':     n,m,q = map(int,input().split())     seq = [c for c in input().split()]     for _ in range(q):         a,b = map(int,input().split())         l = a - 1         r = b - 1         if seq[l] == "<":             print(0)         elif l == r:             if seq[l] not in "<>":                 print(seq[l])             else:                 print(0)         else:             print(main(l,r)) 贴一段代码，不需要每次只减一位字符串的数字，优化了(">"..."<")该种情况下的处理，测试案例都通过了，考试的时候没写完，没法测试

一步步删不行，要看结构，如果一段字符串是封闭的">*****<"，这里面方块的分都可以得完，最后的方向取决于这段里面最大的数。如果不是封闭的就直接跑出界。

我感觉你是不是pop反了，比方说><从左到右，删掉的是>，你好像删掉的是<，题目好像是说（如果当前为》|<且下一个也为这个，那么去掉，我理解的是删掉当前的

等一个大佬，连续的符号怎么处理呀

同，蹲一个大佬来解答一下😂

感觉就按照题目的要求一步一步走就行，也没用什么数据结构 import sys [n,m,q] = list(map(int, sys.stdin.readline().strip().split())) tmp = list(sys.stdin.readline().strip().split()) path = [] for i in tmp:     if i.isdigit():         path.append(int(i))     else:         path.append(i) for _ in range(q):     # 读取每一行     line = sys.stdin.readline().strip()     # 把每一行的数字分隔后转化成int列表     [l, r] = list(map(int, line.split()))     valid = path[l-1: r]     curr = 0     mark = 0     # 1: left, 0: right     orient = 1     # 上一个是'<'或'>'     flag = 0     while curr >= 0 and curr < len(valid):         if valid[curr] == '>':             if flag:                 valid.pop(curr)                 if orient:                     curr = curr                 else:                     curr -= 1             else:                 orient = 1                 curr += 1                 flag = 1         elif valid[curr] == '<':             if flag:                 valid.pop(curr)                 if orient:                     curr = curr                 else:                     curr -= 1             else:                 orient = 0                 curr -= 1                 flag = 1         elif valid[curr] == 0:             valid.pop(curr)             if orient:                 curr = curr             else:                 curr -= 1         else:             mark += valid[curr]             valid[curr] -= 1             flag = 0             if orient:                 curr += 1             else:                 curr -= 1     print(mark)