题目截图

题目分析

想法1，遍历所有kab作为xy的所有可能值，找到其所有因子，然后看看另一个因子是否符合，但这样可能最坏情况k可以从1到1e10直接炸裂
想法2，固定x，然后分解xy的质因数，看看x还差什么，设置还差的东西为base，然后看看base的倍数落在b到d中的值，成立的就作为y即可

想法1：遍历所有可能的xy（tle）

# -*- coding: utf-8 -*-
# @Time    : 2022/10/16 22:22
# @Author  : bridgekiller
# @FileName: E.py
# @Software: PyCharm
# @Blog    ：bridge-killer.blog.csdn.netimport sysinput = sys.stdin.readline# 装饰器
# 解决带有参数和返回值函数的装饰器
# 类似aop的思想, 非侵入式地横切
# --------------------
def debug(func):def wrapper(*args, **kwargs):print('----------------')res = func(*args, **kwargs)print('----------------')return resreturn wrapper# --------------------# 装饰器，解决dfs爆栈问题
# --------------------
# 手写栈模板
# 克服py栈太浅的问题
from types import GeneratorTypedef bootstrap(f, stack=[]):def wrappedfunc(*args, **kwargs):if stack:return f(*args, **kwargs)else:to = f(*args, **kwargs)while True:if type(to) is GeneratorType:stack.append(to)to = next(to)else:stack.pop()if not stack:breakto = stack[-1].send(to)return toreturn wrappedfunc# --------------------
import mathdef factor(num):factors = []for_times = int(math.sqrt(num))for i in range(for_times + 1)[1:]:if num % i == 0:factors.append(i)t = int(num / i)if not t == i:factors.append(t)return factorsdef solve():a, b, c, d = list(map(int, input().split()))# cd >= xy = kabk = (c * d) // (a * b) # [1, k]for mul in range(1, k + 1):xy = mul * a * bfs = factor(xy)for f in fs:if a < f <= c and b < xy // f <= d:return print(f, xy // f)if b < f <= d and a < xy // f <= c:return print(xy // f, f)return print(-1, -1)if __name__ == '__main__':for _ in range(int(input())):solve()

想法2：遍历可能的x（accepted）

# -*- coding: utf-8 -*-
# @Time    : 2022/10/16 22:22
# @Author  : bridgekiller
# @FileName: E.py
# @Software: PyCharm
# @Blog    ：bridge-killer.blog.csdn.netimport sysinput = sys.stdin.readline# 装饰器
# 解决带有参数和返回值函数的装饰器
# 类似aop的思想, 非侵入式地横切
# --------------------
def debug(func):def wrapper(*args, **kwargs):print('----------------')res = func(*args, **kwargs)print('----------------')return resreturn wrapper# --------------------# 装饰器，解决dfs爆栈问题
# --------------------
# 手写栈模板
# 克服py栈太浅的问题
from types import GeneratorTypedef bootstrap(f, stack=[]):def wrappedfunc(*args, **kwargs):if stack:return f(*args, **kwargs)else:to = f(*args, **kwargs)while True:if type(to) is GeneratorType:stack.append(to)to = next(to)else:stack.pop()if not stack:breakto = stack[-1].send(to)return toreturn wrappedfunc# --------------------
import mathdef factor(num):factors = []for_times = int(math.sqrt(num))for i in range(for_times + 1)[1:]:if num % i == 0:factors.append(i)t = int(num / i)if not t == i:factors.append(t)return factorsdef solve():a, b, c, d = list(map(int, input().split()))# cd >= xy = kab# 遍历x# 给ab分解质因数tmp = a * bfactors = [] # [factor, exp]while tmp != 1:for i in range(2, tmp + 1):cnt = 0while tmp % i == 0:cnt += 1tmp //= iif cnt > 0:factors.append([i, cnt])if tmp == 1:break#print(factors)if a * b == 1:return print(a + 1, b + 1)#print(factors)for x in range(a + 1, c + 1):# x = pi ^ ri# ab -> factorsbase = 1for f, needCnt in factors:# 看看x能满足f的多少指数tmp = xcnt = 0while tmp % f == 0:cnt += 1tmp //= fif cnt < needCnt:for _ in range(needCnt - cnt):base *= f#print(f'x:{x}, base:{base}')k1 = b // basek2 = d // basefor k in range(k1, k2 + 1):if b < k * base <= d:return print(x, k * base)return print(-1, -1)if __name__ == '__main__':for _ in range(int(input())):solve()

总结

数论题也要考虑最坏情况的复杂度
有时候有多种遍历方式，不同的解法，要记得计算复杂度

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codeforces：E1. Divisible Numbers (easy version)【数论 + 复杂度计算 + 分解质因数】

目录

题目截图

题目分析

想法1：遍历所有可能的xy（tle）

想法2：遍历可能的x（accepted）

总结

codeforces：E1. Divisible Numbers (easy version)【数论 + 复杂度计算 + 分解质因数】相关推荐

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