http://acm.hdu.edu.cn/showproblem.php?pid=1258 关键点就是一次递归里面一样的数字只能选一次. #include <cstdio> #include <cstring> int n,t; ],c[]; bool flag; void dfs(int k,int sum,int l) { if(sum==t) { ;i<l-;i++) printf("%d+",c[i]); printf(]); flag=; re…

Prime Ring Problem HDU - 1016 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 a…

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…

Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7337    Accepted Submission(s): 4591 Problem Description There is a rectangular room, covered with square tiles. Each tile is color…

#include<cstdio> #include<cstring> #include<cmath> ][]; #define inf 0xffffff int n,m; int min; ]={,,,-}; ]={,,-,}; ][]; int front,rear; void dfs(int x,int y,int c_step) { if(x==n&&y==m) { if(c_step<min) {min=c_step;return;} }…

#include<cstdio> #include<cstring> ]={,,,-}; ]={,,-,}; ][]; int x1,y1,x2,y2; int step; int n,m,t; void dfs(int x,int y,int c_step) { if(x==x2&&y==y2&&c_step==t) { step=;return; } if((x>x2?x-x2:x2-x)+(y>y2?y-y2:y2-y)+c_ste…

很简单的深搜 只要看出来是深搜... 注意判断最后一点是否与加一为质数 #include<stdio.h> #include<string.h> #include<algorithm> #include<map> #include<math.h> using namespace std; int n; int ans[30]; bool vis[30]; bool ok[105]; void init()///素数打表 { memset(ok,t…

Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbe…

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.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1427 思路分析: 题目要求判断是否存在一种运算组合使得4个数的计算结果为24,因为搜索的层次为3层,不需要选择出最短的路径,采用dfs更有效: 拓展状态时,从当前状态拥有的数中选取两个进行某种运算(因为两个数之间存在大小关系,所以对于除法与减法来说,运算顺序一定, 大的数为被减数或被除数:加法与乘法具有交换律,相对顺序没有影响),如果可以进行运算且运算结果满足题目要求,则该状态可以 拓展,如此拓展状…