编程题记录2

二分法

题目链接
https://www.lanqiao.cn/problems/99/learning/?page=1&first_category_id=1&second_category_id=3&tags=%E7%9C%81%E8%B5%9B&tag_relation=intersection

n, k = map(int, input().split())
ls = [[0]*2 for _ in range(n)]
for i in range(n):
    ls[i][0], ls[i][1] = map(int, input().split())
def check(x):
    count = 0
    # 遍历每个长方形
    for j in range(n):
        # 保证最小边长要比x大
        min_len = min(ls[j][0], ls[j][1])
        if min_len < x:
            continue
        else:
            count += (ls[j][0]//x) * (ls[j][1]//x)
    # 如果这种方案能切出的巧克力比人数多，说明边长还能更长
    if count >= k:
        return True
    else:
        return False
# 采用二分法进行遍历
l, r = 0, 100001
while l+1 != r:
    mid = (l+r)//2
    # 如果mid数据可以，说明边长偏小
    if check(mid):
        l = mid
    else:
        r = mid

if check(l):
    print(l)
else:
    print(r)