说明

笔者做的测试的卷子

1

回文串判断，就是给定一个字符串，判断最少修改几个才能变成回文串，就从两边向中间遍历就行

N = int(input())
str_array = list(input())
left,right = 0,N-1
res = 0
while left < right:
    if str_array[left] != str_array[right]:
        res += 1
    left += 1
    right -= 1
print(res)

2.

铺砖快使得相邻的颜色不同，并且满足所有颜色的砖块个数相同

没写出来，只A了0.1，求个大佬帮看看

def calc_euclid(N,M):
    if N % M != 0:
        return calc_euclid(M,N%M)
    return M


T = int(input())
while T:
    T -= 1

    [N,M] = list(map(int,input().split()))
    if N % 2 == 0 or M % 2 == 0:
        print(2)
    elif N == M:
        print(N)
    # elif (N%3==0 and M%3==0) or (M%3==0 and N%3==0):
    #     print(3)
    else:
        val = calc_euclid(N,M)
        print(max(N//val,M//val))

3

拆开一个数字n，使得其拆出来的数字的和能组成[1,n]的所有数字，

第一想法就是不断除以2，因为要使和达到n，至少得有ceil(n/2)才行

from math import ceil
T = int(input())

while T:
    T -= 1
    res = 0
    n = int(input())
    while n > 1:
        divid = ceil(n/2)
        res += 1
        n -= divid
    print(res+1)

4

对一个矩阵，每次取最大行/列，以求出k次的最大和，每次取完后该行/列的值减去d

暴力法，用两个数组存了行和和列和，并找最小的，迭代k次。只A了40.

N,M,k,d = list(map(int,input().split()))
nums = []
row_sum = []
col_sum = []
for _ in range(N):
    val = list(map(int,input().split()))
    nums.append(val)
    row_sum.append(sum(val))

for j in range(M):
    val = 0
    for i in range(N):
        val += nums[i][j]
    col_sum.append(val)

all_sum = 0
while k :
    k-= 1
    max_row_sum = max(row_sum)
    max_col_sum = max(col_sum)
    if  max_row_sum > max_col_sum or (max_row_sum == max_col_sum and N > M):
        row_idx = row_sum.index(max_row_sum)
        all_sum += max_row_sum
        row_sum[row_idx] -= M*d
        for j in range(M):
            col_sum[j] -= d
    else:
        col_idx = col_sum.index(max_col_sum)
        all_sum += max_col_sum
        col_sum[col_idx] -= N*d
        for i in range(N):
            row_sum[i] -= d
print(all_sum)

总结

我好菜啊