简单题. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<queue> #include<map> #include<stack> using namespace std; int main() { long long k,n; scanf("%lld%ll…

#include<stdio.h> #include<iostream> #include<cstdio> #include<queue> #include <vector> #include<stack> #include<cmath> #include<cstring> #include<cstdlib> #include<climits> #include<algorithm…

A. Johny Likes Numbers 题目连接: http://www.codeforces.com/contest/678/problem/A Description Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k. Input The only line con…

[codeforces 55]D. Beautiful numbers 试题描述 Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and…

Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number x is really big if the difference between x and the sum of its digits (in decimal representation) is…

Codeforces Round #597 (Div. 2 Consider the set of all nonnegative integers: 0,1,2,-. Given two integers a and b (1≤a,b≤104). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on. Each number is painted whi…

Alyona and Numbers 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/A Description After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers - the first column containing inte…

http://codeforces.com/problemset/problem/449/D 题意:给n个数,求and起来最后为0的集合方案数有多少 思路:考虑容斥,ans=(-1)^k*num(k),num(k)代表至少有k个数字and起来为1的方案数,那么怎么求num呢? 考虑and起来至少为x的方案数:那么一定是2^y-1,其中y代表有多少个数&x==x,问题就变成有多少数"包含"了某个数(二进制下),用dp解决这个问题:如果某一位数字是1,那么它一定能转移到它不是1的那…

传送:http://codeforces.com/gym/101612 题意:给出一个大小为n的序列a[i],每次选其中一个数乘以一个正整数,问进行k步操作后最少剩下多少种数字,输出0≤k≤n,所有的k的答案. 注意这k步不一定是连续的. 分析: 对于每个数,可以有两种操作: 1. 先将有倍数的数变成它们的最大倍数,而且按照出现次数比较少的先变. 2. 将所有数都变成lcm,而且按照出现次数比较少的先变. 数组f[i]代表,操作i次的最小种类数.对于每一次操作,取min. #include<bi…

[题目链接] http://codeforces.com/problemset/problem/449/D [题目大意] 给出一些数字,问其选出一些数字作or为0的方案数有多少 [题解] 题目等价于给出一些集合,问其交集为空集的方案数, 我们先求交集为S的方案数,记为dp[S],发现处理起来还是比较麻烦, 我们放缩一下条件,求出交集包含S的方案数,记为dp[S], 我们发现dp[S],是以其为子集的方案的高维前缀和, 我们逆序求高维前缀和即可,之后考虑容斥,求出交集为0的情况, 我们发现这个容斥…