题目如下: In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60.  Formally, we want the number of indices i < j with (time[i] + time[j]) % 60…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/ 题目描述 In a list of songs, the i-th song has a duration of time[i] seconds. Return the…
网址:https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/submissions/ 参考:https://blog.csdn.net/Sea_muxixi/article/details/88649191 直接双层for循环必定tle 在一轮遍历中将t取余存入map中 之后只需要遍历一个大小为60的map即可解决问题 class Solution { public: int numPa…
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60.  Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.…
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60.  Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.…
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60.  Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.…
1013. 总持续时间可被 60 整除的歌曲  显示英文描述 我的提交返回竞赛   用户通过次数450 用户尝试次数595 通过次数456 提交次数1236 题目难度Easy 在歌曲列表中,第 i 首歌曲的持续时间为 time[i] 秒. 返回其总持续时间(以秒为单位)可被 60 整除的歌曲对的数量.形式上,我们希望索引的数字  i < j 且有 (time[i] + time[j]) % 60 == 0. 示例 1: 输入:[30,20,150,100,40] 输出:3 解释:这三对的总持续时…
1 在网上找到了对应的网站.其中正在播放的歌曲正是我们想要的 2 在地址栏输入上面音乐网站的网址,并点击捕获广播.(URL直接给出了音乐的完整地址,比如http://www.someserver.com/abc.mp3之类的,则不需要点击捕获广报,直接点击确定即可,但是一般网站是把音乐嵌入到网站中去的,所以需要点击捕获广播,知道这首音乐的真实地址) 3 软件将会自动再打开一个对话框,我在地址栏输入刚才的音乐网站地址,然后点击"转到",则将从你的IE首页跳转到该网站,并且开始播放音乐.…
1012. Complement of Base 10 Integer Every non-negative integer N has a binary representation.  For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on.  Note that except for N = 0, there are no leading z…
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