传送门:http://codeforces.com/contest/899/problem/C 本题是一个数学问题——集合划分. 将集合{1,2,...,n}划分成两个集合,使得两个集合的元素之和的绝对差值最小. 首先,考虑最简单的操作: 第1步:将元素1和n置入集合A: 第2步:将元素2和n-1置入集合B: 第3步:将元素3和n-2置入集合A: …… 第i步:将元素i和n+1-i,当i为奇数时置入集合A,偶数时置入集合B: …… 第n/2步:将元素n/2和n/2+1置入集合B. 如此,集合A中…

A. Splitting in Teams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There were n groups of students which came to write a training contest. A group is either one person who can write the…

C. Dividing the numbers Petya has n integers: 1, 2, 3, ..., n. He wants to split these integers in two non-empty groups in such a way that the absolute difference of sums of integers in each group is as small as possible. Help Petya to split the inte…

[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…

大意: 求将[1,n]划分成两个集合, 且两集合的和的差尽量小. 和/2为偶数最小差一定为0, 和/2为奇数一定为1. 显然可以通过某个前缀和删去一个数得到. #include <iostream> #include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #inclu…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] n为偶数. l = 1, r = n (l,r)放在一组 l++,r-- 新的l,r放在另外一组 直到l+1==r 这个时候,判断两组的和,如果一样的话,分散在两组 差为1否则差为0 n为奇数 l = 2,r = n (l,r)放在一组 l++,r-- 新的l,r放在另外一组 直到l+1==r 这个时候,判断两组的和,如果一样的话,分散在两组 差为0(把1放在那个较少的组) 否则,差为1 1随意放在哪一组都可以 [代码] #in…

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…

http://codeforces.com/problemset/problem/899/C tot为奇数时,绝对差为1:tot为偶数时,绝对差为0. 难点在于如何输出. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<vector> #i…