1400F - x-prime Substrings 首先发现 \(x\) 很小,所以发现对应的 x-prime 字符串数也很少,最多的情况是 \(x = 19\),有 2399 个,先爆搜出来. 现在问题变成了,删去最少的字符,使字符串内没有这些模式串. 那么建 AC 自动机,把模式串终点标记,即在非终点行走,每次可以跳过字符,求跳过的最小值. DP 一下,设 \(f_{i, u}\) 为前 \(i\) 个字符,当前在 AC 自动机上的 \(u\) 号节点,其中没有 x-prime 子串,删除…

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…

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". Now we have another string p. Your job is to find…

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;…

题意:在n个星球,每2个星球之间的联通需要依靠一个网络适配器,每个星球喜欢的网络适配器的价钱不同,先给你一个n,然后n个数,代表第i个星球喜爱的网络适配器的价钱,然后给出一个矩阵M[i][j]代表第i个星球到第j个星球联通所需的价钱,求联通所有星球所需的最小价钱 思路:prime 每次加入一条边的时候把边2个点的权值加进去即可 (zoj崩了,题也没法交啊,不知对错,啊~~~) 代码: #include "iostream" #include "string.h" #…

题意:给你一个矩阵M[i][j]表示i到j的距离 求最小生成树 思路:裸最小生成树 prime就可以了 最小生成树专题 AC代码: #include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set&…