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 (…

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…

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…

题目描述: http://poj.org/problem?id=3414 中文大意: 使用两个锅,盛取定量水. 两个锅的容量和目标水量由用户输入. 允许的操作有:灌满锅.倒光锅内的水.一个锅中的水倒入另一个锅. 在两个锅互相倒水的过程中,若一个锅被倒满,而原来的锅中还有水,则剩余在原来的锅中. 不断执行这三个过程,直到任意一个锅中,贮存了目标数量的水,过程结束. 思路: 队列节点记录的是当前两个锅的贮水量和之前的一系列操作. 在弹出队列首节点,获取了当前两个锅的水量信息后,后续怎么操作,有 6…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…