题意:给定两个数 m,n,问你在从1到 n,和从 1到 m中任选两个数加起来是5的倍数,问你有多少个. 析:先计算 m 和 n中有多少个取模5是从0到4的,然后根据排列组合,相乘就得到了小于等于 m 和 n的并且能整除五的个数,然后再加上剩下的. 代码如下: #include <bits/stdc++.h> using namespace std; typedef long long LL; const int aa = 1234567; const int bb = 123456; cons…
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…
A. Alyona and Numbers 题目连接: http://www.codeforces.com/contest/682/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 integers…
A. Home Numbers 题目连接: http://www.codeforces.com/contest/638/problem/A Description The main street of Berland is a straight line with n houses built along it (n is an even number). The houses are located at both sides of the street. The houses with od…
题目链接: B. Modulo Sum time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a sequence of numbers a1, a2, ..., an, and a number m. Check if it is possible to choose a non-empty subse…
Problem G.Giant ScreenTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86821#problem/G Description You are working in Advanced Computer Monitors (ACM), Inc. The company is building and selling giant c…
题目链接: A. Round House time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entra…
A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<math.h> #include<vector> using namespace…
这题一眼看就是水题,map随便计 然后我之所以发这个题解,是因为我用了log2()这个函数判断在哪一层 我只能说我真是太傻逼了,这个函数以前听人说有精度问题,还慢,为了图快用的,没想到被坑惨了,以后尽量不用 #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #…
题意:给定 n 和 k,然后是 n 个数,表示1-k的一个值,问你修改最少的数,使得所有的1-k的数目都等于n/k. 析:水题,只要用每个数减去n/k,然后取模,加起来除以2,就ok了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cma…