题目简述:给定字符串$s[1 \dots n](n \leq 2 \times 10^5)$,以及$Q \leq 2 \times 10^5$个询问,每个询问有两个参数$1 \leq l \leq r \leq n$,求 $$ \sum_{i=l}^r \operatorname{lcp}(s[l \dots r], s[i \dots r]), $$ 其中$\operatorname{lcp}(s, t)$表示字符串$s$和$t$的最长公共前缀(Longest Common Prefix)的长…

题目链接  Tricky Function $f(i, j) = (i - j)^{2} + (s[i] - s[j])^{2}$ 把$(i, s[i])$塞到平面直角坐标系里,于是转化成了平面最近点对问题. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >=…

D. Tricky Function time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important cont…

题目链接:点击打开链接 暴力出奇迹. 正解应该是近期点对.以i点为x轴,sum[i](前缀和)为y轴,求随意两点间的距离. 先来个科学的暴力代码: #include<stdio.h> #include<string.h> #include<vector> #include<algorithm> #include<iostream> #include<queue> using namespace std; #define N 10005…

现在我们在工作中,在开发中都会或多或少的用到图表统计数据显示给用户.通过图表可以很直观的,直接的将数据呈现出来.这里我就介绍说一下利用百度开源的echarts图表技术实现的具体功能. 1.对于不太理解echarts是个怎样技术的开发者来说,可以到echarts官网进行学习了解,官网有详细的API文档和实例供大家参考学习. 2.以下是我在工作中实现整理出来的实例源码: 公用的支持js文件 echarts.js.echarts.min.js,还有其他的图表需要支持的js文件也可以到官网下载 echa…

http://codeforces.com/contest/678/problem/D D. Iterated Linear Function Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, nand x find the value of g(n)(x) mod…

M. Heaviside Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/problem/M Description Heaviside function is defined as the piecewise constant function whose value is zero for negative argument and one for non-negat…

D. Tricky Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 codeforces.com/problemset/problem/429/D Description Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The sele…

A. Calculating Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem/A Description For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn Your task is to calculate f(n) for a…

http://codeforces.com/problemset/problem/837/E   题意: f(a, 0) = 0; f(a, b) = 1 + f(a, b - gcd(a, b)) 输出f(a,b) a=A*gcd(a,b)    b=B*gcd(a,b) 一次递归后,变成了 f(A*gcd(a,b),(B-1)*gcd(a,b)) 若gcd(A,(B-1))=1,那么 这一层递归的gcd(a,b)仍等于上一层递归的gcd(a,b) 也就是说,b-gcd(a,b),有大量的时间…