Prime Words A prime number is a number that has only two divisors: itself and the number one. Examples of primenumbers are: 1, 2, 3, 5, 17, 101 and 10007.In this problem you should read a set of words, each word is composed only by letters in the ran…

UVA - 524 Prime Ring Problem Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  into each circle separately, and the sum of number…

CF912E Prime Gift题解(搜索+二分答案) 标签:题解 阅读体验:https://zybuluo.com/Junlier/note/1314956 洛谷题目链接 $     $ CF题目链接 先翻译一下题意 给你\(n\)个质数,让你求这\(n\)个质数能乘起来组成的第\(K\)大数(\(n\le 16,K\)好大八大) 思路 看到这个数据范围,很容易想到搜索对吧 但是又不能盲目地搜索,考虑\(meet-in-the-middle\) 搜出前\(8\)个和后\(8\)个能拼出的所有…

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…

题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1141 题意:判断区间[a,b]的f(i)是否为素数,f(i)=i*i+i+40: 思路:打个表,然后注意精度: #include<bits/stdc++.h> using namespace std; #define ll long long #define esp…

题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1141 题意: 给定区间,求区间内所有整数a,f(a) = a * a + a - 1为质数的概率. 分析: 卡精度卡的蛋疼.. 最后要加个eps处理四舍五入问题..不是很懂.. 代码: #include<cstdio> const int maxm = 1e4 +…

  Prime Ring Problem  A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always…

题目链接:Uva 552 思路分析:时间限制为3s,数据较小,使用深度搜索查找所有的解. 代码如下: #include <iostream> #include <string.h> using namespace std; ; int n; int A[MAX_N], vis[MAX_N]; int is_prime( int n ) { ) return true; ; i * i <= n; ++i ) ) return false; return true; } voi…

10140 - Prime Distance 题目链接 题意:求[l,r]区间内近期和最远的素数对. 思路:素数打表,打到sqrt(Max)就可以,然后利用大的表去筛素数.因为[l, r]最多100W.所以能够去遍历一遍.找出答案. 注意1的情况,一開始没推断1,结果WA了 代码: #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define INF 0x…

题意  把1到n这n个数以1为首位围成一圈  输出全部满足随意相邻两数之和均为素数的全部排列 直接枚举排列看是否符合肯定会超时的  n最大为16  利用回溯法 边生成边推断  就要快非常多了 #include<cstdio> using namespace std; const int N = 50; int p[N], vis[N], a[N], n; int isPrime(int k) { for(int i = 2; i * i <= k; ++i) if(k % i == 0)…