数位dp解决的问题是指求在一段数的区间里面 满足条件的数的个数 核心为两点 http://wenku.baidu.com/link?url=tpfIYzhx_MzevpIM58UZ66pr-87MCFPKTMKFdGDi5jUqyO9ckti0mY6diSz2PZEL_ZBhd2zIbhus1mnzDiAO1B5K2Vu38YDsqjmOvYKFT6q 我自己总结下吧 第一 是dp的状态转移方程 dp[i][j] 表示的是以j开头的i位数 满足条件的个数为多少 既然是要记录满足条件的个 那么 d…

1.HDU 2089  不要62    简单数位dp 2.总结:看了题解才敲出来的,还是好弱.. #include<iostream> #include<cstring> #include<cstdio> using namespace std; ][]; //dp[i][j]表示以j为首位符合条件的i位数的个数 void init() //预处理 { memset(dp,,sizeof(dp)); dp[][]=; ;i<;i++){ ;j<;j++){…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5898 题意:很明确,找出区间[l , r]中符合连续奇数为偶数,连续偶数为奇数的个数. 思路:dp[i][j][1]表示i位数j开头符合条件的数,dp[i][j][0]表示i位数j开头(之后)可能符合条件的数. #include<cstdio> #include<cstring> using namespace std; typedef long long ll; ll dp[22][…

[HDU  3555]原题直通车: 代码: // 31MS 900K 909 B G++ #include<iostream> #include<cstdio> #include<cstring> using namespace std; int a[20]; __int64 dp[20][11]; void digit_dp() { memset(dp, 0LL, sizeof(dp)); dp[0][0]=1; for(int i=1; i<20; ++i)…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=2089 不要62 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 34901    Accepted Submission(s): 12664 Problem Description 杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoe…

Bomb HDU - 3555 (数位DP) 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 the sub-sequence "49&…

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…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4734 F(x) Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4389    Accepted Submission(s): 1614 Problem Description For a decimal number x with…

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…

题意: 我们定义十进制数x的权值为f(x) = a(n)*2^(n-1)+a(n-1)*2(n-2)+...a(2)*2+a(1)*1,a(i)表示十进制数x中第i位的数字. 题目给出a,b,求出0~b有多少个不大于f(a)的数. 思路: 数位DP,用来学习数位DP了. <数位DP> 所谓数位DP就是基于考虑数字的每一位来转移的DP. 例如求比456小的数,可以这么考虑, 4          5               6   4       5             (0~6) 4…