题目链接:点这 github链接:(包含数据和代码,题解):点这 链接:https://www.nowcoder.com/acm/contest/104/E来源:牛客网 题目描述 (受限于评测机,此题数据范围与现场赛不一致,请谅解) Once upon a time, there was a beautiful princess named TQM, and a handsome prince named GSS. One day, the prince would like to visit…

题目描述:链接点此 这套题的github地址(里面包含了数据,题解,现场排名):点此 链接:https://www.nowcoder.com/acm/contest/104/D来源:牛客网 题目描述 Do you remember Kanna-chan we met last year? She is so cute, and this year, she entered middle school, with Cirno. As we know, Cirno is bad at math, s…

如下图这是“今日头条杯”首届湖北省大学程序设计竞赛的第一题,作为赛后补题 题目描述:链接点此 这套题的github地址(里面包含了数据,题解,现场排名):点此 Let  be a regualr triangle, and D is a point in the triangle. Given the angle of . Then let AD, CD and BD form a new triangle, what is the size of the three angles? 输入描述:…

题目描述:链接点此 这套题的github地址(里面包含了数据,题解,现场排名):点此 题目描述 Given n positive integers , your task is to calculate the product of these integers, The answer is less than 输入描述: The first line of input is an integer n, the i-th of the following n lines contains the…

题目描述:链接点此 这套题的github地址(里面包含了数据,题解,现场排名):点此 题目描述 由于系统限制,C题无法在此评测,此题为现场赛的D题 Who killed Cock Robin? I, said the Sparrow, With my bow and arrow,I killed Cock Robin. Who saw him die? I, said the Fly.With my little eye,I saw him die. Who caught his blood?…

题目描述:链接点此 这套题的github地址(里面包含了数据,题解,现场排名):点此 链接:https://www.nowcoder.com/acm/contest/104/H来源:牛客网 题目描述 Mingming, a cute girl of ACM/ICPC team of Wuhan University, is alone since graduate from high school. Last year, she used a program to match boys and…

链接:https://www.nowcoder.com/acm/contest/104/G来源:牛客网 题目描述 Given n positive integers , your task is to calculate the product of these integers, The answer is less than 题解:直接python高精度 坑:c++高精度会T 紫书上的高精度乘法改不来 t = int(input()) p=1 for i in range(t): s = i…

链接:https://ac.nowcoder.com/acm/contest/283/H来源:牛客网 题目描述 由于临近广西大学建校90周年校庆,西大开始了喜闻乐见的校园修缮工程! 然后问题出现了,西大内部有许许多多的道路,据统计有N栋楼和M条道路(单向),每条路都有“不整洁度”W,现在校方想知道从S楼到T楼的所有路径中,“不整洁度”乘积最小是多少. 由于答案可能很大,所以你需要将最后的答案对109+7取模. 输入描述: 第一行为四个整数N.M.S.T,意义如上. 第2至第M+1行每行表示一条道…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2441    Accepted Submission(s): 1415 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4331    Accepted Submission(s): 2603 Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) =…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 155 Accepted Submission(s): 110   Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x.If…

题目 A Simple Math Problem 解析 矩阵快速幂模板题 构造矩阵 \[\begin{bmatrix}a_0&a_1&a_2&a_3&a_4&a_5&a_6&a_7&a_8&a_9\\ 1&0&0&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0&0&0\\ 0&…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2791    Accepted Submission(s): 1659 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

P2071 -- A Simple Math Problem IX 时间限制:1000MS      内存限制:262144KB 状态:Accepted      标签:    数学问题-博弈论   无   无 Description 给定a,b,n,保证a≥2,b≥1,a^b≤n.两个人在玩游戏,每个人每次可以把a加1,或者把b加1,但是不能违反a^b<=n,无法再进行操作的人就输掉了这一场游戏. 假设两个人都足够聪明,按照最优策略进行游戏,问先手是否有必胜策略. Input Format 第…

A Simple Math Problem [题目链接]A Simple Math Problem [题目类型]矩阵快速幂 &题解: 这是一个模板题,也算是入门了吧. 推荐一个博客:点这里 跟着这个刷,应该就可以了 &代码: #include <cstdio> #include <iostream> #include <set> #include <cmath> #include <cstring> #include <al…

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1697    Accepted Submission(s): 959 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =…

A Simple Math Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1645    Accepted Submission(s): 468 Problem Description Given two positive integers a and b,find suitable X and Y to meet th…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=1757 A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6621    Accepted Submission(s): 4071 Problem Description Lele now is thin…

A Simple Math Problem 一个矩阵快速幂水题,关键在于如何构造矩阵.做过一些很裸的矩阵快速幂,比如斐波那契的变形,这个题就类似那种构造.比赛的时候手残把矩阵相乘的一个j写成了i,调试了好久才发现.改过来1A. 贴个AC的代码: const int N=1e5+10; ll k,m,s[10]; struct mat { ll a[10][10]; }; mat mat_mul(mat A,mat B) { mat res; memset(res.a,0,sizeof(res.a…

A Simple Math Problem Lele now is thinking about a simple function f(x).If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 .Now, I will give a0 ~ a9 and two posit…

2018 ACM-ICPC 中国大学生程序设计竞赛线上赛:https://www.jisuanke.com/contest/1227 题目链接:https://nanti.jisuanke.com/t/26219 Rock Paper Scissors Lizard Spock Description: Didi is a curious baby. One day, she finds a curious game, which named Rock Paper Scissors Lizard…

题 Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive…

Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 .Now, I will give a0 ~ a9 and two positiv…

http://acm.hdu.edu.cn/showproblem.php?pid=1757 Problem Description Lele now is thinking about a simple function f(x).If x < 10  f(x) = x.If x >= 10  f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);And ai(0<=i<=9) can only be…

Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive in…

Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 .Now, I will give a0 ~ a9 and two positiv…

Problem Description Lele now is thinking about a simple function f(x). If x < f(x) = x. If x >= f(x) = a0 * f(x-) + a1 * f(x-) + a2 * f(x-) + …… + a9 * f(x-); And ai(<=i<=) can only be or . Now, I will give a0 ~ a9 and two positive integers k…

题目地址:HDU 1757 最终会构造矩阵了.事实上也不难,仅仅怪自己笨..= =! f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10) 构造的矩阵是:(我代码中构造的矩阵跟这个正好是上下颠倒过来了) |0 1 0 ......... 0|    |f0|   |f1 | |0 0 1 0 ...... 0|    |f1|   |f2 | |...................1| *  |..| = |...…

Lele now is thinking about a simple function f(x).  If x < 10 f(x) = x.  If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);  And ai(0<=i<=9) can only be 0 or 1 .  Now, I will give a0 ~ a9 and two positive integers k…

Lele now is thinking about a simple function f(x).  If x < 10 f(x) = x.  If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);  And ai(0<=i<=9) can only be 0 or 1 .  Now, I will give a0 ~ a9 and two positive integers k…