Simulation volume leetcode [general] 1894 Find the student number who needs to add chalk

1894. Find the number of students who need to add chalk

There are... In one class  n  A student , The number is 0  To n - 1 . Each student will answer the questions in turn , The number is 0  First of all, I want to answer , And then it's numbered 1  Of the students , And so on , Until the number is n - 1  Of the students , Then the teacher will repeat the process , Renumber from 0  The students began to answer questions .

Give you a length of n  And subscript from 0  The starting array of integers  chalk  And an integer  k . In the beginning, there were a total of  k  A piece of chalk . When the number is  i  When students answer questions , He will consume chalk[i]  A piece of chalk . If there's any chalk left Strictly less than  chalk[i] , So students i  need Add   chalk .

Please return to need Add   Chalk students Number  .

Example 1：

Input ：chalk = [5,1,5], k = 22
Output ：0
explain ： Students' consumption of chalk is as follows ：

• The number is 0 The students who use 5 A piece of chalk , then k = 17 .
• The number is 1 The students who use 1 A piece of chalk , then k = 16 .
• The number is 2 The students who use 5 A piece of chalk , then k = 11 .
• The number is 0 The students who use 5 A piece of chalk , then k = 6 .
• The number is 1 The students who use 1 A piece of chalk , then k = 5 .
• The number is 2 The students who use 5 A piece of chalk , then k = 0 .
  The number is 0 Of the students don't have enough chalk , So he needs chalk .
  Example 2：

Input ：chalk = [3,4,1,2], k = 25
Output ：1
explain ： Students' consumption of chalk is as follows ：

• The number is 0 The students who use 3 A piece of chalk , then k = 22 .
• The number is 1 The students who use 4 A piece of chalk , then k = 18 .
• The number is 2 The students who use 1 A piece of chalk , then k = 17 .
• The number is 3 The students who use 2 A piece of chalk , then k = 15 .
• The number is 0 The students who use 3 A piece of chalk , then k = 12 .
• The number is 1 The students who use 4 A piece of chalk , then k = 8 .
• The number is 2 The students who use 1 A piece of chalk , then k = 7 .
• The number is 3 The students who use 2 A piece of chalk , then k = 5 .
• The number is 0 The students who use 3 A piece of chalk , then k = 2 .
  The number is 1 Of the students don't have enough chalk , So he needs chalk .

Tips ：

chalk.length == n
1 <= n <= 105
1 <= chalk[i] <= 105
1 <= k <= 109

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/find-the-student-that-will-replace-the-chalk
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Code ：

import time
from typing import List


class Solution:
    def __init__(self):
        pass

    def chalkReplacer(self, chalk: List[int], k: int) -> int:
        k = k%sum(chalk)
        for i,c in enumerate(chalk):
            k-=c
            if k<0:return i


def test(data_test):
    s = Solution()
    return s.chalkReplacer(*data_test)


def test_obj(data_test):
    result = [None]
    obj = Solution(*data_test[1][0])
    for fun, data in zip(data_test[0][1::], data_test[1][1::]):
        if data:
            res = obj.__getattribute__(fun)(*data)
        else:
            res = obj.__getattribute__(fun)()
        result.append(res)
    return result


if __name__ == '__main__':
    datas = [
        [[5,1,5],22],
        [[3,4,1,2],25],
        [[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],999999999],
    ]
    for data_test in datas:
        t0 = time.time()
        print('-' * 50)
        print('input:', data_test)
        print('output:', test(data_test))
        print(f'use time:{
      time.time() - t0}s')

remarks ：
GitHub：https://github.com/monijuan/leetcode_python

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