Bomb Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 16713 Accepted Submission(s): 6128 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorists impr…

http://acm.hdu.edu.cn/showproblem.php?pid=4507 Problem Description 单身! 依旧单身! 吉哥依旧单身! DS级码农吉哥依旧单身! 所以.他生平最恨情人节,无论是214还是77.他都讨厌! 吉哥观察了214和77这两个数,发现: 2+1+4=7 7+7=7*2 77=7*11 终于,他发现原来这一切归根究竟都是由于和7有关!所以,他如今甚至讨厌一切和7有关的数. 什么样的数和7有关呢? 假设一个整数符合以下3个条件之中的一个.那么我…

XHXJ's LIS Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2265    Accepted Submission(s): 927 Problem Description #define xhxj (Xin Hang senior sister(学姐)) If you do not know xhxj, then careful…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3555 Bomb Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) 问题描述 The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bo…

给出n,问所有[0,n]区间内的数中,不含有49的数的个数 数位dp,记忆化搜索 dfs(int pos,bool pre,bool flag,bool e) pos:当前要枚举的位置 pre:当前要枚举的位置的前面是否为4 flag:枚举当前时,这个数的49时候被算过了 e:当前位置是否可以随便取值 dp[pos][pre][flag] #include<cstdio> #include<cstring> #include<algorithm> #include<…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555 Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 18739    Accepted Submission(s): 6929 Problem Description The counter-terrorists found…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 7921    Accepted Submission(s): 2778 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorists…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555 简单的数位DP入门题目 思路和hdu2089基本一样 直接贴代码了,代码里有详细的注释 代码: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> using namespace std; ][]; void init() { dp[][]=;dp[][]=;dp[…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6156 题意:如题. 解法:数位DP,暴力枚举进制之后,就转化成了求L,R区间的回文数的个数,这个直接做一个数位DP就好了.dp[jz][start][cur][state]表示jz进制下以start位起始到cur位状态为state(1表示已经回文,0表示没有回文)时回文数的个数. #include <bits/stdc++.h> using namespace std; typedef long…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 19122    Accepted Submission(s): 7068 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorist…