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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…
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…
Description Math Olympiad is called “Aoshu” in China. Aoshu is very popular in elementary schools. Nowadays, Aoshu is getting more and more difficult. Here is a classic Aoshu problem: ABBDE __ ABCCC = BDBDE In the equation above, a letter stands for…
Description Math Olympiad is called “Aoshu” in China. Aoshu is very popular in elementary schools. Nowadays, Aoshu is getting more and more difficult. Here is a classic Aoshu problem: ABBDE __ ABCCC = BDBDEIn the equation above, a letter stands for a…
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…
A very hard Aoshu proble 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 prob…
题目链接:https://vjudge.net/problem/UVALive-5107 题目大意:用ABCDE代表不同的数字,给出形如ABBDE___ABCCC = BDBDE的东西: 空格里面可以填入+-*/的运算符,给字母赋予不同的值,问有多少种情况使得 等式成立. 题目分析: 可以直接用大模拟+暴力求解,注意对于重复情况的判重. 给出代码: #include <iostream> #include <set> #include <algorithm> #incl…
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)…
题意:题意:给你3个字符串s1,s2,s3;要求对三个字符串中的字符赋值(同样的字符串进行同样的数字替换), 替换后的三个数进行四则运算要满足左边等于右边.求有几种解法. Sample Input 2 A A A BCD BCD B   Sample Output 5 72 eg:ABBDE   ABCCC   BDBDE :令 A = 1, B = 2, C = 0, D = 4, E = 5 12245 + 12000 = 24245: 注意没有前导零! ! #include<stdio.h…
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…