2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 D Thinking-Bear magic (几何) 链接:https://ac.nowcoder.com/acm/contest/163/D来源:牛客网 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 In order to become a magical girl, Thinking-Bear are learning m…

链接:https://www.nowcoder.com/acm/contest/163/F 来源:牛客网 2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 F Color it 时间限制:C/C++ 3秒,其他语言6秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 There is a matrix A that has N rows and M columns. Each grid (i,j)(0 ≤ i < N,…

2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 H - A Simple Problem with Integers (线段树,循环节) 链接:https://ac.nowcoder.com/acm/contest/163/H 来源:牛客网 链接:https://ac.nowcoder.com/acm/contest/163/H来源:牛客网 时间限制:C/C++ 2秒,其他语言4秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %ll…

2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 F Color it (扫描线) 链接:https://ac.nowcoder.com/acm/contest/163/F来源:牛客网 时间限制:C/C++ 3秒,其他语言6秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 There is a matrix A that has N rows and M columns. Each grid (i,j)(0 ≤ i…

2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 J Beautiful Numbers (数位DP) 链接:https://ac.nowcoder.com/acm/contest/163/J?&headNav=acm来源:牛客网 时间限制:C/C++ 8秒,其他语言16秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 NIBGNAUK is an odd boy and his taste is strange a…

题目链接:Fruit Ninja 比赛链接:2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛 题目描述 Fruit Ninja is a juicy action game enjoyed by millions of players around the world, with squishy, splat and satisfying fruit carnage! Become the ultimate bringer of sweet, tasty destruction wit…

Rabbits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1193    Accepted Submission(s): 628 Problem Description Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number li…

整理代码... Little Boxes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2304    Accepted Submission(s): 818 Problem Description Little boxes on the hillside.Little boxes made of ticky-tacky.Littl…

今天做的沈阳站重现赛,自己还是太水,只做出两道签到题,另外两道看懂题意了,但是也没能做出来. 1. Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description ACM ICPC is launching a thic…

题目链接:https://ac.nowcoder.com/acm/contest/163/J 题目大意:给定一个数N,求区间[1,N]中满足可以整除它各个数位之和的数的个数.(1 ≤ N ≤ 1012). 输入: 21018 输出: Case 1: 10Case 2: 12 解题思路:比较简单的一道数位dp题,因为N的范围最大可为10的十二次方,即数位和的范围为[1,108],1-108的最小公倍数很大不可求,所以我们直接暴力枚举数位和为1-108的情况,然后利用数位dp求出合法数的个数就可以了…

A链接:https://www.nowcoder.com/acm/contest/163/A Fruit Ninja is a juicy action game enjoyed by millions of players around the world, with squishy, splat and satisfying fruit carnage! Become the ultimate bringer of sweet, tasty destruction with every sl…

题目描述 We consider a positive integer perfect, if and only if it is equal to the sum of its positive divisors less than itself. For example, 6 is perfect because 6 = 1 + 2 + 3. Could you write a program to determine if a given number is perfect or not?…

题目描述 In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field, or, more generally, in a ring or even a semiring. The matrix product is designed for representing the…

题意:求小于等于N且能被自己所有位上数之和整除的数的个数. 分析:裸的数位dp.用一个三位数组dp[i][j][k]记录:第i位,之前数位之和为j,对某个mod余数为k的状态下满足条件的个数.这里mod的值就是小于等于N的数中,所有可能出现的数位之和.所以solve函数中需要对dfs函数做一个循环,上限是9*pos(数位之和不会超过9*pos). 还有需要注意的是,在递归的时候可以通过判断当前数位之和sum是否超过mod,超过的话肯定在这个状态下没有满足条件的数,以此剪枝优化. #include…

1001题意:n个人,给m对敌对关系,X个好人,Y个坏人.现在问你是否每个人都是要么是好人,要么是坏人. 先看看与X,Y个人有联通的人是否有矛盾,没有矛盾的话咋就继续遍历那些不确定的人关系,随便取一个数3,与其相连的就是4,间隔就要相同,dfs搜过去就可以判断了 #include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define…

C.Recursive sequence 求ans(x),ans(1)=a,ans(2)=b,ans(n)=ans(n-2)*2+ans(n-1)+n^4 如果直接就去解...很难,毕竟不是那种可以直接化成矩阵的格式,我们也因为这个被卡很长时间 事实上可以把这道式子化成几个基本元素的格式,然后就容易组合了,比如ans(n-2)*2+ans(n-1)+(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^1+1 包含了所有的基本组成形式,化绝对为相对,并且除了一个n-2其他都是n…

题目链接:http://acm.hdu.edu.cn/search.php?field=problem&key=2016ACM%2FICPC%D1%C7%D6%DE%C7%F8%B4%F3%C1%AC%D5%BE-%D6%D8%CF%D6%C8%FC%A3%A8%B8%D0%D0%BB%B4%F3%C1%AC%BA%A3%CA%C2%B4%F3%D1%A7%A3%A9&source=1&searchmode=source A.染色乱搞. #include <bits/stdc…

废话: 这道题很是花了我一番功夫.首先,我不会kmp算法,还专门学了一下这个算法.其次,即使会用kmp,但是如果暴力枚举的话,还是毫无疑问会爆掉.因此在dfs的基础上加上两次剪枝解决了这道题. 题意: 我没有读题,只是队友给我解释了题意,然后我根据题意写的题. 大概意思是给n个字符串,从上到下依次标记为1——n,寻找一个标记最大的串,要求这个串满足:标记比它小的串中至少有一个不是它的子串. 输入: 第一行输入一个整型t,表示共有t组数据. 每组数据首行一个整型n,表示有n个串. 接下来n行,每行…

HDU 6222 Heron and His Triangle 链接:http://acm.hdu.edu.cn/showproblem.php?pid=6222 思路: 打表找规律+大数运算 首先我们可以打个表暴力跑出前几个满足题意的T 打表代码: #include<bits/stdc++.h> using namespace std; bool fun(double n) { if(abs(round(n) - n) < 0.000000000000001) ; ; } int ma…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=5979 按AC顺序: I - Convex Time limit    1000 ms Memory limit 65536 kB OS Windows We have a special convex that all points have the same distance to origin point. As you know we can get N segments after linki…

Little Boxes Problem Description Little boxes on the hillside.Little boxes made of ticky-tacky.Little boxes.Little boxes.Little boxes all the same.There are a green boxes, and b pink boxes.And c blue boxes and d yellow boxes.And they are all made out…

n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two of them (labelled aand b, where 1≤a≠b≤n) withstood the test of time. Two monks, Yuwgna and Iaka, decide to make…

题意:有\(n\)个点,\(m\)个集合,集合\(E_i\)中的点都与集合中的其它点有一条边权为\(t_i\)的边,现在问第\(1\)个点和第\(n\)个点到某个点的路径最短,输出最短路径和目标点,如果不满足条件则输出\(Evil John\). 题解:题目所给的边数关系太复杂了,我们可以让每个集合中的所有点都与一个虚拟节点连边,而这些点两两却不连,然后再去找\(1\)个和第\(n\)个点的最短路径,不难发现,最终得到的路径为\(dis[i]/2\),所以我们只要用虚拟节点建边然后跑两次dijk…

题意:给你\(n\)个字符串,\(s_1,s_2,...,s_n\),对于\(i(1\le i\le n)\),找到最大的\(i\),并且满足\(s_j(1\le j<i)\)不是\(s_i\)的子串. 题解:直接\(O(n^2)\)然后跑kmp匹配,这里注意要剪枝,不然会T,也就是说对于前\(i-1\)个串,如果它是后面某个串的子串,那么我们就不用对它跑kmp,因为它后面的某个串包含了它. 代码: int t; int n; string s[N]; int ne[N]; bool st[N]…

题意:有\(n\)个数,开始给你两个数\(a\)和\(b\),每次找一个没出现过的数\(i\),要求满足\(i=j+k\)或\(i=j-k\),当某个人没有数可以选的时候判他输,问谁赢. 题解:对于\(a\)和\(b\),我们能有他两得到的最小数一定是\(d=gcd(a,b)\),所以总共能选的数的个数为\(n/d\),判断奇偶即可. 代码: int t; int n,a,b; int main() { ios::sync_with_stdio(false);cin.tie(0);cout.ti…

Tree Ming and Hong are playing a simple game called nim game. They have nn piles of stones numbered 11 to nn ,the ii-th pile of stones has a_iai​ stones. There are n - 1n−1 bidirectional roads in total. For any two piles, there is a unique path from…

Travel There are nn planets in the MOT galaxy, and each planet has a unique number from 1 \sim n1∼n. Each planet is connected to other planets through some transmission channels. There are mm transmission channels in the galaxy. Each transmission cha…

Swap There is a sequence of numbers of length nn, and each number in the sequence is different. There are two operations: Swap the first half and the last half of the sequence (if nn is odd, the middle number does not change) Swap all the numbers in…

Angel's Journey “Miyane!” This day Hana asks Miyako for help again. Hana plays the part of angel on the stage show of the cultural festival, and she is going to look for her human friend, Hinata. So she must find the shortest path to Hinata’s house.…

Tasks It's too late now, but you still have too much work to do. There are nn tasks on your list. The ii-th task costs you t_iti​seconds. You want to go to bed TT seconds later. During the TT seconds, you can choose some tasks to do in order to finis…