python实现洛谷P1319 压缩技术

把每一行的0或1用bool值表示

List = list(map(int, input().split()))
Count = 0
k = 1
Bool = False #因为矩阵内是0与1，所以我认为用bool类型方便点
while k < len(List):
    for i in range(0, List[k]):  #List[k]存放0或1的连续次数
        print(int(Bool), end = '')
        Count += 1
        if Count % List[0] == 0:
            print()
    k += 1
    Bool = not Bool #输出完该次bool值的List[k]次，则转换bool值

个人因看错题又尝试了矩阵转置输出······

```python
List = list(map(int, input().split()))
array = []#存放所有行，形成二维数组
Bool = False
k = 1
array1 = []#中间变量，存放每一行的bool值
Count = 0
while k < len(List):
    for i in range(0, List[k]):
        array1.append(int(Bool))#把每一个bool值放入行内
        Count += 1
        if Count % List[0] == 0:
            array.append(array1)#若该行已满，则该行放入列表内
            array1 = []
    k += 1
    Bool = not Bool
Count = 0
for i in range(0, List[0]):#矩阵的转置输出
    for j in range(0, List[0]):
        print(array[j][i], end = '')
        Count += 1
        if Count % List[0] == 0:
            print()