题目链接:https://vjudge.net/problem/38405 #include<bits/stdc++.h> using namespace std; ][]; ]; long long dfs(int pos,int preok,int pre1) { ) ; ) return dp[pos][pre1]; :b[pos]; ; ; ;i<=up;i++) { ,,i==); ,,i==); } if (preok) dp[pos][pre1]=ans; return a…

/** 题目:poj3208 Apocalypse Someday 链接:http://poj.org/problem?id=3208 题意:求第K(K <= 5*107)个有连续3个6的数. 思路:数位dp+二分. dp[i][j]表示长度为i,前缀状态为j时含有的个数. j=0表示含有前导0: j=1表示前缀连续1个6 j=2表示前缀连续2个6 j=3表示前缀连续3个6 j=4表示前缀不是6: */ //#include<bits/stdc++.h> #include<cstr…

题目链接: http://hihocoder.com/problemset/problem/1301?sid=804672 题解: 二分答案,每次判断用数位dp做. #include<iostream> #include<cstring> #include<cstdio> using namespace std; typedef long long LL; const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; //dp[x][0]表示高位还没有出…

积累点: 1: (l&r)+((l^r)>>) == (l+r)/2 2: 注意判断现在是否有限制.当枚举下一个量时,是(isQuery && j==end),不要搞错. 传送门:http://acm.upc.edu.cn/problem.php?id=2223 题意: 能被7整除或者含7的数称为A-Number,所有A-Number从小到大写好,下标编号(从1开始),去掉那些下标为A-Number的数,剩下的数称为B-Number.求第N个B-Number是多少. 思…

All submissions for this problem are available. Chef likes numbers and number theory, we all know that. There are N digit strings that he particularly likes. He likes them so much that he defines some numbers to be beautiful numbers based on these di…

思路:dp[i][j]:表示第i位在B进制下数字和. 用二分找第k个数! 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 using namespace std; ][],…

SNIBB Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1485    Accepted Submission(s): 435 Problem Description   As we know, some numbers have interesting property. For example, any even number h…

K-th Nya Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 3155    Accepted Submission(s): 1021 Problem Description Arcueid likes nya number very much.A nya number is the number which has…

数位DP加二分 //数位dp,dfs记忆化搜索 #include<iostream> #include<cstdio> #include<cstring> using namespace std; typedef long long LL; #define N 20 LL dp[N][3];//dp[i][j]表示长度为i,前面有j个6时不含666的数的个数 int num[N]; //c6表示前面6的个数 LL dfs(int len, int c6, bool is…

Solution 我好像写了一个非常有趣的解法233, 我们可以用数位$DP$ 算出比$N$小的数中 字典序比 $X$ 小的数有多少个, 再和 $rank$进行比较. 由于具有单调性, 显然可以二分答案. Code #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; ll m, rank, sum[][][]; ], len, a…