Intuitive one to learn about Grundy basic :) Now every pile becomes a game, so we need to use Sprague-Grundy Theory. Calculation is quite intuitive - and if you print them out, you will find these Grundy numbers loops by 9. def firstMissing(s): ret =…

https://www.hackerrank.com/contests/w2/challenges/manasa-and-stones 简单题. #include<iostream> using namespace std; int main() { int T; cin >> T; while (T--) { int n, a, b; cin >> n >> a >> b; // a < b if ( a > b) { int tm…

Change language : Manasa 和 她的朋友出去徒步旅行.她发现一条小河里边顺序排列着带有数值的石头.她开始沿河而走,发现相邻两个石头上的数值增加 a 或者 b. 这条小河的尽头有一个宝藏,如果Manasa能够猜出来最后一颗石头上的数值,那么宝藏就是她的.假设第一个石头的上数值为0,找出最后一个石头的可能的所有数值. 输入格式 第一行包含整数 T, 代表测试数据的组数. 每组数组包含三行: 第一行包含 n,代表石头的个数 第二行包含 a 第三行包含 b 输出格式 升序输出最后一…

original version hackerrank programming version 题目大意是定义了一个正整数的表,第一行是1,第二行是1,2,第三行1,2,3...定义prime triple是在表上八连通的三个质数.然后问某行有多少个质数至少在一个prime triple中.   行数 <= 1e7. 题解: 假设要求第n行的答案,只要把上2行和下2行的数都抠出来,然后判断一下就好了.问题转化为求[L,R]之间的质数,这里5行数,区间长度大概有5e7,所以要用modified 区…

Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. Credits:Special thanks to @mithmatt for adding this problem and creating all test cases. 求n以内的所有素数,以前看过的一道题目,通过将所有非素数标记出来,再找出素数,代码如下: public i…

Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers! Input The input begins with the number t of test cases in a single line (t<=10). In each of the next t li…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

传送门 Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1, 2, . . . , n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle sho…

题意为给出两个四位素数A.B,每次只能对A的某一位数字进行修改,使它成为另一个四位的素数,问最少经过多少操作,能使A变到B.可以直接进行BFS搜索 #include<bits/stdc++.h> using namespace std; bool isPrime(int n){//素数判断 || n == ) return true; else{ ; ; i < k; i++){ ) return false; } return true; } } ]; ]; void getPrime…

hdu5901题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5901 code vs 3223题目链接:http://codevs.cn/problem/3223/ 思路:主要是用了一个Meisell-Lehmer算法模板,复杂度O(n^(2/3)).讲道理,我不是很懂(瞎说什么大实话....),下面输出请自己改 #include<bits/stdc++.h> using namespace std; typedef long long LL;…