//hdoj 3555 //2013-06-27-16.53 #include <stdio.h> #include <string.h> __int64 dp[21][3], n; int len, bit[21]; //dp[i][0] 长度为i 包含49的个数 //dp[i][1] 长度为i没有49但以9开头的 //dp[i][2] 长度为i 没有49 void init() { dp[0][2] = 1; for (int i = 1; i < 20; i++) {…

数位DP的DFS写法.... Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4630    Accepted Submission(s): 1614 Problem Description The counter-terrorists found a time bomb in the dust. But this time…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555 思路分析:该问题要求求解1—N中的数中含有49的数的个数,可以使用DFA来递推dp公式:详细解释点击链接查看: 代码如下: #include <cstdio> #include <cstring> #include <iostream> using namespace std; + ; ]; int digit[MAX_N]; int NextState(int cu…

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…

hdu 3555 Bomb 题意: 在1~N(1<=N<=2^63-1)范围内找出含有 ‘49’的数的个数: 与hdu 2089 不要62的区别:2089是找不不含 '4'和 '62'的区间范围内的数,此题是含有:正好相反,对于 "不要62"只是用第二位表示首位数字,这一题呢? 看转化:易知一定要要知道首位是9的个数,才能在前面加4得到 '49',但是什么状态能从不含 '49'转移到含 '49'?直接在不含'49'前加9,那么就出现了三个状态之间的递推转化,从而推出了第二维…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 10419    Accepted Submission(s): 3673 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorist…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the curre…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555 题目大意:从0开始到给定的数字N所有的数字中遇到“49”的数字的个数. Sample Input 3 1 50 500   Sample Output 0 1 15 Hint From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","24…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 11804    Accepted Submission(s): 4212 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorist…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 15102    Accepted Submission(s): 5452 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorists…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 7926    Accepted Submission(s): 2780 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorists…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 6609    Accepted Submission(s): 2303 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 题意:上一题是不要62 这个是"不要49" 代码: #include <stdio.h> #include <ctime> #include <math.h> #include <limits.h> #include <complex> #include <string> #include <functio…

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…

链接: https://vjudge.net/problem/HDU-3555 题意: The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes…

Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 24148    Accepted Submission(s): 9092 Problem Description The counter-terrorists found a time bomb in the dust. But this time the terrorist…

RT. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ],ret=,dp[][][]; void get_bit(long long x) { ret=; ;x/=;} } long long dfs(long long pos,long long pre,bool flag1,long long flag2) {…

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<…

题意 求区间[1,n]内含有相邻49的数. 思路 比较简单的按位DP思路.这是第一次学习记忆化搜索式的数位DP,确实比递推形式的更好理解呐,而且也更通用~可以一般化: [数位DP模板总结] int dfs(int pos, int pre, int flag, bool limit) { if (pos == -1) return flag==target_flag; if (!limit && ~dp[pos][pre][flag]) return dp[pos][pre][flag];…

转载请注明出处:http://blog.csdn.net/a1dark 分析:初学数位DP完全搞不懂.很多时候都是自己花大量时间去找规律.记得上次网络赛有道数位DP.硬是找规律给A了.那时候完全不知数位DP为何物.不过还是有很多时候要用数位DP.比如当一个数字超过了数组承受的极限.不能再打表AC.先看这道题.首先划分状态.然后初始化.最后从高位向低位状态转移.代码含详解 //dp[len][0]表示长度为len不含49的数量 //dp[len][1]表示长度为len不含44但以9开头的数字的数量…

题意: 有N个炸弹..每个炸弹有两个位置可以选择..把炸弹放到其中一个地方去...炸弹的爆炸范围是其为圆心的圆...两个炸弹不能有攻击范围上的重合..问要满足条件..炸弹爆炸范围的半径最长能是多少... 题解: 每个炸弹看成一类..其在两个中比选一个..符合2-sat的构图条件....那么就二分枚举炸弹的爆炸范围..枚举相互是否干扰来做边构造2-sat模型...tarjan来判断是否合法.. 题目不难..但是我2B了...在初始化vector时..for (i=0;i<(n<<1);i+…

题意: 给一个数字n,求从1~n中有多少个数是含有49的,比如49,149,1490等都是含49的. 思路: 2^64也顶多是十进制的20多位,那么按十进制位来分析更简单.如果能计算k位十进制数中分别有多少个含49的,那么计算就简单了. 首先要求关于十进制位的一些信息,比如:i位的十进制数包含49有多少个,不包含49的多少个(除掉最高位是9的数量),不包含49但是最高位是9的有多少个(因为可能和更高一位组合成49).注意精度,注意爆longlong. #include <bits/stdc++.…

数位dp: 数位dp是一种计数用的dp,一般就是要统计一个区间[li,ri]内满足一些条件数的个数.所谓数位dp,字面意思就是在数位上进行dp.数位的含义:一个数有个位.十位.百位.千位......数的每一位就是数位.一般都通过dfs来递归找寻. 数位dp与dfs爆搜之间的区别就是前者会记录状态以便优化下一次寻找的时间(也就是记忆化操作) 数位dp操作: 例如给你一个区间[12,12345],你需要找到这个区间内满足条件的数的个数.这个时候我们一般都是先找[0,12345]这个区间满足条件数的个…

数位DP cxlove基础数位DP第二题 与上题基本相同(其实除了变成long long以外其实更简单了……) //HDOJ 3555 #include<cmath> #include<vector> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for…

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): 7316    Accepted Submission(s): 2551 Problem Description The counter-terrorists found a time…

题目链接:http://acm.split.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): 15372    Accepted Submission(s): 5563 Problem Description The counter-terrorists f…

HDU 2089 不要62 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2089 Problem Description 杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer).杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可以消除个别的士司机和乘客的心理障碍,更安全地服务大众.不吉利的数字为所有含有4或62的号码.例如:62315 73418 88914都属于不吉利号码.但是,6115…

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…

 数位DP总结 By Wine93 2013.7 1.学习链接 [数位DP] Step by Step   http://blog.csdn.net/dslovemz/article/details/8540340 [总结] 数位统计模板 http://www.cnblogs.com/jffifa/archive/2012/08/17/2644847.html 2.各人总结领悟 数位DP关键在于状态的设计,设计状态时要考虑当前状态(如dp[pos][s1][s2]),一旦当前状态确定后,pos后…

        ID Origin Title   62 / 175 Problem A CodeForces 55D Beautiful numbers   30 / 84 Problem B HDU 4352 XHXJ's LIS   108 / 195 Problem C HDU 2089 不要62   89 / 222 Problem D HDU 3555 Bomb   59 / 107 Problem E POJ 3252 Round Numbers   47 / 75 Problem…