Codeforces 55D Beautiful Number a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. Input The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two…

题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2829 题目描述: Mike is very lucky, as he has two beautiful numbers, 3 and 5. But he is so greedy that he wants infinite beautiful numbers. So he declares that any positive number which is…

题目链接: http://codeforces.com/problemset/problem/55/D 数位DP 题目描述: 一个数能被它每位上的数字整除(0除外),那么它就是beautiful number.问区间[a,b]上有多少个beautiful number.如102就是一个beautiful number,因为它能整除1,2.14不是,因为14不能整除4. 解法: 数位DP,设dp[i][j][k]为累计到第i为,公倍数为j,模lcm(1,2,```,9)=2520的余数为k的数的个…

Beautiful Number Time Limit: 2 Seconds      Memory Limit: 65536 KB Mike is very lucky, as he has two beautiful numbers, 3 and 5. But he is so greedy that he wants infinite beautiful numbers. So he declares that any positive number which is dividable…

beautiful number 问题描述 令 A = \sum_{i=1}^{n}a_i * {10}^{n-i}(1\leq a_i \leq 9)A=∑​i=1​n​​a​i​​∗10​n−i​​(1≤a​i​​≤9)(nn为AA的位数).若AA为“漂亮的数”当且仅当对于任意1 \leq i < n1≤i<n满足a[i] \geq a[i+1]a[i]≥a[i+1]且对于任意1 \leq i \leq n,i < j \leq n1≤i≤n,i<j≤n,满足a[i]a[i]…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2829 Beautiful Number Time Limit: 2 Seconds      Memory Limit: 65536 KB Mike is very lucky, as he has two beautiful numbers, 3 and 5. But he is so greedy that he wants infinite beautiful nu…

beautiful number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 801    Accepted Submission(s): 518 Problem Description Let A=∑ni=1ai∗10n−i(1≤ai≤9)(n is the number of A's digits). We call A as “…

把数位dp写成记忆化搜索的形式,方法很赞,代码量少了很多. 下面为转载内容:  a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits.    问一个区间内[l,r]有多少个Beautiful数字    范围9*10^18        数位统计问题,构造状态也挺难的,我想不出,我的思维局限在用递推去初始化状态,而这里的状态定义也比较难    跟pre的…

题意:给你两个数p和x,然后让你找出一个长度为p的数,把它的最后移到最前面之后得到的数是原来数字的x倍,有很多这样的数取最小. 思路:枚举最后一位,然后就可以推出整个的一个数,然后比较得到的数的第一个数字和枚举的数字是否相等既可以. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; int p,x; ]; int ma…

DFS. /* 5179 */ #include <iostream> #include <algorithm> #include <map> #include <cstdio> #include <cstring> using namespace std; #define MAXN 1500 map<int, bool> tb; int a[MAXN], n, v; ][] = { {,}, {,,,}, {,,,,}, {,,,}…