A very hard Aoshu problem Problem Description Aoshu is very popular among primary school students. It is mathematics, but much harder than ordinary mathematics for primary school students. Teacher Liu is an Aoshu teacher. He just comes out with a pro…

http://acm.hdu.edu.cn/showproblem.php?pid=4403 题意: 给出一串数字,在里面添加一个等号和多个+号,使得等式成立,问有多少种不同的式子. 思路: 数据量比较小,直接dfs爆搜答案即可. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<map> using namespace std;…

题意:给你一个数字字符串.问在字符串中间加'='.'+'使得'='左右两边相等. 1212  : 1+2=1+2,   12=12. 12345666 : 12+3+45+6=66.  1+2+3+4+56=66: #include<stdio.h> #include<string.h> #include<string> #include<map> #include<stack> #include<math.h> #include&l…

暴力$dfs$. 先看数据范围,字符串最长只有$15$,也就是说枚举每个字符后面是否放置“$+$”号的复杂度为${2^{15}}$. 每次枚举到一种情况,看哪些位置能放“$=$”号,每个位置都试一下,然后判断一下是否可行. 最坏复杂度$O({2^{15}}*{15^2})$,事实上是达不到最坏复杂度的. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<c…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=4403 A very hard Aoshu problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1080    Accepted Submission(s): 742 Problem Description Aoshu is ver…

A very hard Aoshu problem                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Aoshu is very popular among primary school students. It is mathematics, but much harder than o…

HASH+暴力. /* 4403 */ #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <map> using namespace std; #define MAXN 55 map<]; char s[MAXN]; int get(int b, int e) { ; for (int i=b; i<e; ++i)…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5135 题目大意:给你n条边,选出若干条边,组成若干个三角形,使得面积和最大.输出最大的面积和. 先将边从小到大排序,这样前面的两条边加起来如果不大于第三条边就可以跳出,这是一个存在性条件. dfs(int idx,int now,int cnt,int nowmax)代表我当前处理的是第idx条边,已经加入边集的有cnt条边,当前的边的长度和为now,组成的最大面积和为nowmax. 暴力枚举每个三…

Problem Description You are given N sets.The i−th set has Ai numbers.You should divide the sets into L parts.And each part should have at least one number in common.If there is at least one solution,print YES,otherwise print NO.   Input In the first…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4770 思路:由于最多只有15个".",可以直接枚举放置的位置,然后判断是否能够全部点亮即可.需要注意的是,有一个特殊的light,也需要枚举它的位置以及放置的方向. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using names…