目录 @description@ @solution@ @accepted code@ @details@ @description@ 已知 a 序列,并给定以下关系: \[\begin{cases} f(1, j) = a_j & (1 \le j \le n) \\ f(i, j) = \min\{f(i - 1, j), f(i - 1, j - 1)\} + a_j & (2 \le i \le j \le n) \end{cases}\] 给定 m 次询问 (xi, yi),求…

题目:http://codeforces.com/problemset/problem/455/E 题意:给定数组a,及f的定义: f[1][j] = a[j];  1 <= j <= n f[i][j] = min(f[i-1][j-1], f[i-1][j]) + a[j];  2 <= i <= j <= n 给定q个询问,每个询问为l,r,求f[l][r] My solution: 写一些小数据就可以发现,其实对于一个询问l,r,其实等价于: 从[r-l+1, r]这…

1. CF 438D The Child and Sequence 大意: n元素序列, m个操作: 1,询问区间和. 2,区间对m取模. 3,单点修改 维护最大值, 取模时暴力对所有>m的数取模. 因为取模后至少减半, 复杂度$O(nlognlogC)$ 2. CF 431E Chemistry Experiment 大意: n个试管, 第$i$个试管有$a_i$单位水银, m个操作: 1, 修改$a_x$改为$v$. 2, 将$v$单位水倒入试管, 求一种方案使得有水的试管水银与水总量的最大…

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…

Vasya is studying number theory. He has denoted a function f(a, b) such that: f(a, 0) = 0; f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b. Vasya has two numbers x and y, and he wants to calculate f(x, y).…