http://www.codeforces.com/contest/477/problem/C 题目大意:给你n个集合,每个集合里面有四个数字,他们的gcd是k,输出符合条件的集合中m,m为集合中最大的数,且保证m要尽量小. 思路:由找规律可以得到集合的关系为1+6*k  2+6*k  3+6*k  5+6*k. 不过我的写法不是这样...(道理还是一样的) //看看会不会爆int!数组会不会少了一维! //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> u…

Codeforces Round #272 (Div. 2) A. Dreamoon and Stairs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move.…

D. Dreamoon and Sets 题目连接: http://www.codeforces.com/contest/476/problem/D Description Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both a and b. Let S be a set of exactly four distinct inte…

http://codeforces.com/contest/476/problem/C C. Dreamoon and Sums time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output Dreamoon loves summing up something for no reason. One day he obtains two…

题目链接 题意: 1-m中,四个数凑成一组,满足任意2个数的gcd=k,求一个最小的m使得凑成n组解.并输出 分析: 直接粘一下两个很有意思的分析.. 分析1: 那我们就弄成每组数字都互质,然后全体乘以k不就行了么…… 然后看了看样例…… 这个该怎么说……我是觉得额这道题的output暴露了数据规律怎么破……我算是看出规律再证明的方式A的这道题 当时我看到22那个样例的时候……在想他干嘛要把22放这里……然后发现 2/4/6/10 14/16/18/22也是行的哇…… 化成乘以k之前的数据………

题目链接 这个题取模的时候挺坑的!!! 题意:div(x , b) / mod(x , b) = k( 1 <= k <= a).求x的和 分析: 我们知道mod(x % b)的取值范围为 1  - (b-1).那么我们可以从这一点入口来进行解题.. mod (x, b) = 1 时, x  =  b + 1, 2b + 1, 3b + 1..... a * b + 1. mod (x , b) = 2 时, x =  2b + 2, 4b + 2, 6b + 2, ..... 2a * b…

A. Dreamoon and Stairs 题意:给出n层楼梯,m,一次能够上1层或者2层楼梯,问在所有的上楼需要的步数中是否存在m的倍数 找出范围,即为最大步数为n(一次上一级),最小步数为n/2+n%2 在这个范围里找即可 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<algorithm> using namespace std…

A. Dreamoon and Stairs time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of mo…

按照题意构造集合即可 注意无解情况的判断 #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <map> using namespace std; int n,sum; int main(){ scanf("%d",&n); ){ printf("…

E. Dreamoon and Strings 题目连接: http://www.codeforces.com/contest/476/problem/E Description Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates that is defined…