What doesn't kill you makes you stronger. 那些没有彻底击败你的东西只会让你更强大. Where there is life, there is hope, and the hope simulates us to strive for making it true. Someone told me that most people live such a kind of life: youth is a blunder, manhood a strugg…

A friend without faults will never be found. 没有缺点的朋友是永远找不到的. You can't find a friends without faults, and you can't find a lover without faults. In personal communication, a tolerant heart is very important because it can help you build the relations…

If you keep on believing, the dreams that you wish will come true. 如果你坚定信念,就能梦想成真. I always believe I am unique and I will get what I dream of. But they haven't come true yet. Maybe my efforts are not enough. Please fling yourself into your missions…

Every has the capital to dream. 每个人都有做梦的本钱. Your vision, our mission. That is an advertisment of UMoney Baidu, a financial app from Baidu. It seems that personal financial service has gradually become the most profitable market, and a lot of companie…

I figure life is a gift and I don't intend on wasting it. 我觉得生命是一份礼物,我不想浪费它. Tonight when I was running, I heard two boys said that if we are still alive, there will be hope, I smiled then laughed in my heart, maybe my smile was just mocking, because…

Knowing yourself is the height of wisdom. 了解自己就是大智慧. Two-day holiday, even I didn't have enought time to know about her, I could feel I am not the type she loves. I ever thought I might be not so bad a man. Maybe I was wrong. So, I choose to give up.…

有一些草,一开始高度都是0,它们的生长速率不同. 给你一些单增的日期,在这些日期要将>b的草的部分都割掉,问你每次割掉的部分有多少. 将草的生长速率从大到小排序,这样每次割掉的是一个后缀,而且不会影响它们生长速率的递增性. 就是三种操作,一种对一个后缀赋值,一种对整个数组作 + 另一个数组(d(i)-d(i-1))*a,一种求区间和. 可以通过打标记的线段树实现,标记下放通过预处理生长速率数组的前缀和可以实现. 队友的代码: #include <iostream> #include &l…

平面上给你n(不超过2000)个点,问你能构成多少个面积在[A,B]之间的Rt三角形. 枚举每个点作为直角顶点,对其他点极角排序,同方向的按长度排序,然后依次枚举每个向量,与其对应的另一条直角边是单调的,可以用一个pointer做出来,然后可以得出那些同方向的向量的区间(这个代码好像有点问题,可能会退化,最好确定了一个LL之后,对一个方向的不要重复算RR.这里如果也改成二分就比较好,复杂度不会退化).然后通过二分可以得到A使得面积在[A,B]间的有哪些(其实这个因为也是单调的,好像也没必要二分,…

f(n)定义为n的十进制表示下所有位的平方和. 问你方程K*f(n)=n在a<=n<=b中的解的个数. 发现f(n)最大不超过2000,可以直接枚举f(n),然后判断K*f(n)的位的平方和是否恰好为f(n). #include<cstdio> #include<iostream> using namespace std; typedef long long ll; ll K,a,b; int main(){ // freopen("g.in",&q…

No matter how resourceful you are, you can't fight fate. 人纵有万般能耐,终也敌不过天命. I find that I gradually become a fatalist, because I feel that I have little control over my life. Many times I feel that no matter what I do, there is little change in the con…