Gym - 100338E 题意:给你n,k问在1-n中能整出k的字典序最小的数.范围1018 思路:比较简单的贪心了,枚举10的幂m,然后加上k-m%k, 更新答案就可以了,数据一定要用unsigned long long,我就在这里挂了几次,查了半天. #include <iostream> #include <cstdio> #include <fstream> #include <algorithm> #include <cmath> #…

题目:http://codeforces.com/gym/100338/attachments 贪心,每次枚举10的i次幂,除k后取余数r在用k-r补在10的幂上作为候选答案. #include<bits/stdc++.h> using namespace std; typedef unsigned long long ull; ; ull base[maxbit], n, k; void preDeal() { ] = ; ; i < maxbit; i++){ *]; } } voi…

题目链接 http://codeforces.com/gym/101102/problem/J Description standard input/output You are given an array A of integers of size N, and Q queries. For each query, you will be given a set of distinct integers S and two integers L and R that represent a…

Shopping 题目连接: http://codeforces.com/gym/100803/attachments Description Your friend will enjoy shopping. She will walk through a mall along a straight street, where N individual shops (numbered from 1 to N) are aligned at regular intervals. Each shop…

题目链接:http://codeforces.com/gym/101775/problem/B Aori is very careless so she is always making troubles. One day she does it again, with N big troubles! But this time she seems to be at ease because she has found M Inklings to take all the blames. Eac…

Pavel is developing another game. To do that, he again needs functions available in a third-party library too famous to be called. There are mm functions numbered from 11to mm, and it is known that the ii-th version of the library contains functions…

(-￣▽￣)-* 这道题涉及高精度除法,模板如下: ]; ];//存储进行高精度除法的数据 bool bignum_div(int x) { ,num=; ;s[i];i++) { num=num*+s[i]-'; division[tot++]=num/x+'; num%=x; } division[tot]='\0';//利于进行strcpy() ) //有适合的除数 { ; ') i++; strcpy(s,division+i);//比如49->07,那么下一轮s就变成7,多余的i个0都…

题目链接 \(Description\) 给定\(n\)个数,每次可以将任意一个数乘上任意一个正整数. 求\(k\)次操作后,数列中数的种类最少可以是多少.对每个\(0\leq k\leq n\)输出答案. \(n\leq 3\times10^5,a_i\leq10^6\). \(Solution\) 有两种贪心策略,一是每次找一个出现次数最少的数,把它变成所有数的LCM:二是找一个出现次数最少,且它的某个倍数存在于原数列里的数,将它变成它的这个倍数. 对两种情况取\(\min\)即可. //9…

题意:有个邮递员,要送信,每次最多带 m 封信,有 n 个地方要去送,每个地方有x 封要送,每次都到信全送完了,再回去,对于每个地方,可以送多次直到送够 x 封为止. 析:一个很简单的贪心,就是先送最远的,如果送完最远的还剩下,那么就送次远的,如果不够了,那么就加上回来的距离,重新带够 m 封信,对于左半轴和右半轴都是独立的,两次计算就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include &l…

After the data structures exam, students lined up in the cafeteria to have a drink and chat about how much they have enjoyed the exam and how good their professors are. Since it was late in the evening, the cashier has already closed the cash registe…