4740. 跑圈
摘要
Title: 4740. 跑圈
Tag: 模拟
Memory Limit: 64 MB
Time Limit: 1000 ms
Powered by:NEFU AB-IN
4740. 跑圈
-
题意
见原题
-
思路
分类讨论
- 当方向与上次相同时：直接加上总路程，算出加的圈数
- 当方向与上次不同时：先判断是否经过起点，若经过，则需变方向，那么总路程减去此次路程之后，就需要取绝对值（因为方向变了），然后按照上面的情况处理
-
代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61'''
Author: NEFU AB-IN
Date: 2023-07-09 00:47:29
FilePath: \Acwing\4740\4740.py
LastEditTime: 2023-07-09 01:55:01
'''
# import
from sys import setrecursionlimit, stdin, stdout, exit
from collections import Counter, deque
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from bisect import bisect_left, bisect_right
from datetime import datetime, timedelta
from string import ascii_lowercase, ascii_uppercase
from math import log, gcd, sqrt, fabs, ceil, floor
class sa:
def __init__(self, x, y):
self.x = x
self.y = y
def __lt__(self, a):
return self.x < a.x
# Final
N = int(1e5 + 10)
M = 20
INF = int(2e9)
# Define
setrecursionlimit(INF)
input = lambda: stdin.readline().rstrip("\r\n") # Remove when Mutiple data
read = lambda: map(int, input().split())
AR = lambda x=0: [x] * N
# —————————————————————Division line ——————————————————————
t, = read()
for _ in range(t):
L, cnt, C = 0, 0, 0
l, n = read()
for i in range(n):
d, c = input().split()
d = int(d)
if i == 0:
C = c
if c == C:
L += d
cnt += L // l
else:
if L >= 0 and L - d < 0:
C = c
L -= d
if L < 0:
L = -L
cnt += L // l
L %= l
print(f"Case #{_ + 1}: {cnt}")