HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS(3)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4135 Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1176    Accepted Submission(s): 427 Problem Description Given a number N, you are a…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42600    Accepted Submission(s): 18885 Problem Description A ring is compose of n circles as shown in diagram. Put natural num…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices…

在刚刚写完代码的时候才发现我以前交过这道题,可是没有过. 后来因为不理解代码,于是也就不了了之了. 可说呢,那时的我哪知道什么DFS深搜的东西啊,而且对递归的理解也很肤浅. 这道题应该算HDU 2610 Sequence one的简化版,判重也非常简单. 其他也没有什么好说的了,直接上代码吧. //#define LOCAL #include <iostream> #include <cstdio> #include <cstring> #include <cma…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description A ring is compose of n circles as shown in diagram. Put natural number 1, 2…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1973 Prime Path Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description The ministers of the cabinet were quite upset by the message from the Chief of Secu…

Problem Description A ring is compose of n circles as shown in diagram. Put natural number 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 should always be 1…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1016 Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63806    Accepted Submission(s): 27457 Problem Description A ring is compos…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18313    Accepted Submission(s): 8197 Problem Description A ring is compose of n circles as shown in diagram. Put natural numbe…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12105    Accepted Submission(s): 5497 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…

题目链接 题意 : 就是把n个数安排在环上,要求每两个相邻的数之和一定是素数,第一个数一定是1.输出所有可能的排列. 思路 : 先打个素数表.然后循环去搜..... #include <cstdio> #include <cstring> #include <iostream> using namespace std ; ]; ] ,cs[]; int n ; void get_prime() { memset(prime,,sizeof(prime)); prime[…

按照题意的要求逐渐求解: #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ]={,,,,,,,,}; ][]={","2*3","2*3*5","2*3*5*7","2*3*5*7*11","2*3*5*7*11*13", "2*3*5*7*11…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31031    Accepted Submission(s): 13755 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34846    Accepted Submission(s): 15441 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…

题意:给你2个数n m.从n变成m最少须要改变多少次. 当中: 1.n  m  都是4位数 2.每次仅仅能改变n的一个位数(个位.十位.百位.千位),且每次改变后后的新数为素数 思路:搜索的变形题,这次我们要搜得方向是改变位数中的一位,然后往下搜.直到求出我们须要的那个解 #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<cmath> u…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20105    Accepted Submission(s): 9001 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…

先写一遍思路,跟素数表很类似吧. 1)从小到大遍历数据范围内的所有数.把包含质因子的数的位置都设成跟质因子的位置相同. 2)同一个数的位置可能被多次复写.但是由于是从小到大遍历,这就保证了最后一次写入的是该数的最大质因子的位置 一道题墨迹了好久,上代码分析 #include <iostream> #include <stdio.h> #include <math.h> #include <algorithm> #include <string.h>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 题目大意:输入一个n,环从一开始到n,相邻两个数相加为素数. #include <iostream> #include <cstdio> #include <cstring> using namespace std; ],q[],n; int sushu(int n) { int i; ; i*i<=n; i++) ) ; ; } void dfs(int x,…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 67252    Accepted Submission(s): 28829 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…

Prime Path Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 987    Accepted Submission(s): 635 Problem Description The ministers of the cabinet were quite upset by the message from the Chief of S…

Prime Ring Problem A ring is compose of n circles as shown in diagram. Put natural number 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 should always be 1.…

http://acm.hdu.edu.cn/showproblem.php?pid=1195 这道题虽然只是从四个数到四个数,但是状态很多,开始一直不知道怎么下手,关键就是如何划分这些状态,确保每一个状态都能遍历到. 得到四个数之后,分三种情况处理,每次改变一个数之后都要加入队列,最先输出的就是步数最少. #include <cstdio> #include <cstring> #include <queue> using namespace std; struct p…

题意: 给一个N.然后给M个数,问1~N-1里面有多少个数能被这M个数中一个或多个数整除. 思路: 首先要N-- 然后对于每一个数M 事实上1~N-1内能被其整除的 就是有(N-1)/M[i]个 可是会出现反复 比方 例子 6就会被反复算 这时候我们就须要容斥原理了 加上一个数的减去两个数的.. 这里要注意了 两个数以上的时候 是求LCM而不是简单的相乘! 代码: #include "stdio.h" #include "string.h" #include &qu…

Prime Bases Problem Description Given any integer base b >= 2, it is well known that every positive integer n can be uniquely represented in base b. That is, we can write n = a0 + a1*b + a2*b*b + a3*b*b*b + ... where the coefficients a0, a1, a2, a3,…

Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20927    Accepted Submission(s): 8023 Problem Description There are N villages, which are numbered from 1 to N, and you should…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34799    Accepted Submission(s): 15411 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…

Besides the ordinary Boy Friend and Girl Friend, here we define a more academic kind of friend: Prime Friend. We call a nonnegative integer A is the integer B’s Prime Friend when the sum of A and B is a prime. So an integer has many prime friends, fo…

类似于素数打表. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 1000100 using namespace std; int f[maxn]; void inti() { ; ; i<maxn; i++) { ) { num++; for(int j=i; j<maxn; j+=i) { f[j]=num; } } } } int main() { int n…

一切见凝视. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; bool vis[22]; int n; int ans[22]; int top; bool isprime(int x)//推断素数 { for(int i=2;i<x;i++) if(x%i==0)return false; return…