链接: http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1005&cid=595 若一个整数的个位数字截去,再从余下的数中,减去个位数的2倍,如果差是7的倍数,则原数能被7整除.如果一次不容易看出,就需要继续上述过程.如6139,过程如下:613-9×2＝595,59-5×2＝49,所以6139是7的倍数. 能被11整除的数的特征把一个数由右边向左边数,将奇位上的数字与偶位上的数字分别加起来,再求它们的差,如果这个差是11的倍…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5373 The shortest problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 995    Accepted Submission(s): 498 Problem Description In this problem, w…

The shortest problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description In this problem, we should solve an interesting game. At first, we have an…

The shortest problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 346    Accepted Submission(s): 167 Problem Description In this problem, we should solve an interesting game. At first, we hav…

The shortest problem http://acm.hdu.edu.cn/showproblem.php?pid=5373 Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 181 Accepted Submission(s): 92 Problem Description In this problem, we should sol…

The shortest problem Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 969 Accepted Submission(s): 491 Problem Description In this problem, we should solve an interesting game. At first, we have an…

2018 HDU多校第四场赛后补题 自己学校出的毒瘤场..吃枣药丸 hdu中的题号是6332 - 6343. K. Expression in Memories 题意: 判断一个简化版的算术表达式是否合法. 题解: 注意细节即可. 代码: #include <bits/stdc++.h> using namespace std; int n; char s[505]; int main () { int T; cin>>T; for ( ; T; --T) { scanf(&quo…

2018 HDU多校第三场赛后补题 从易到难来写吧,其中题意有些直接摘了Claris的,数据范围是就不标了. 如果需要可以去hdu题库里找.题号是6319 - 6331. L. Visual Cube 题意: 在画布上画一个三维立方体. 题解: 模拟即可. 代码: #include <bits/stdc++.h> using namespace std; int a, b, c, R, C; char g[505][505]; int main () { int T; cin >>…

2015 HDU 多校联赛 5363 Key Set 题目: http://acm.hdu.edu.cn/showproblem.php? pid=5363 依据前面给出的样例,得出求解公式 fn = 2^(n-1) - 1, 数据量大,实际就是求幂次方. 可用分治法求解.复杂度O(nlogn) // 分治法求高速幂 #include <bits/stdc++.h> using namespace std; typedef unsigned long long ull; const int M…

2015 HDU 多校联赛 5317 RGCDQ 筛法求解 题目  http://acm.hdu.edu.cn/showproblem.php? pid=5317 本题的数据量非常大,測试样例多.数据量大, 所以必须做预处理.也就是用筛法求出全部的F[x],将全部F[x] 打印出来发现.事实上结果不大,最大的数值是7.所以对于每一个区间询问, 直接暴力求取有多少个 1 2 3 4 5 6 7 就可以,从大到小查找.假设出现2个以上 3-7 的数值,那么最大公约数就是该数字. 假设没有出现两个反复…