题目传送门 /* 贪心+构造:因为是对称的,可以全都左一半考虑,过程很简单,但是能想到就很难了 */ /************************************************ Author :Running_Time Created Time :2015-8-3 9:14:02 File Name :B.cpp *************************************************/ #include <cstdio> #include &…

Palindrome Transformation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless…

C. Palindrome Transformation     Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arro…

门户:Codeforces Round #277 (Div. 2) 486A. Calculating Function 裸公式= = #include <cstdio> #include <cstring> #include <algorithm> using namespace std ; typedef long long LL ; LL n ; int main () { while ( ~scanf ( "%I64d" , &n )…

Codeforces Round #277 (Div. 2) A. Calculating Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)n…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:在n^n的海洋里是否有k块陆地 构造算法:按奇偶性来判断,k小于等于所有点数的一半,交叉输出L/S 输出完k个L后,之后全部输出S:) 5 10 的例子可以是这样的: LSLSL SLSLS LSLSL SLSLS SSSSS */ #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> usi…

题目传送门 /* 贪心/数学:还以为是BFS,其实x1 + 4 * k = x2, y1 + 4 * l = y2 */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; char s[MAXN][MAXN]; int main(void) //Codeforces Round #212 (Div. 2)…

题目传送门 /* 题意:n个数字转盘,刚开始每个转盘指向一个数字(0~n-1,逆时针排序),然后每一次转动,奇数的+1,偶数的-1,问多少次使第i个数字转盘指向i-1 构造:先求出使第1个指向0要多少步,按照这个次数之后的能否满足要求 题目读的好累:( */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath>…

题目传送门 /* 题意:给出一系列读者出行的记录,+表示一个读者进入,-表示一个读者离开,可能之前已经有读者在图书馆 构造:now记录当前图书馆人数,sz记录最小的容量,in数组标记进去的读者,分情况讨论一下 */ /************************************************ * Author :Running_Time * Created Time :2015-8-6 0:23:37 * File Name :B.cpp *****************…

题目传送门 /* 题意:删除若干行,使得n行字符串成递增排序 暴力+构造:从前往后枚举列,当之前的顺序已经正确时,之后就不用考虑了,这样删列最小 */ /************************************************ Author :Running_Time Created Time :2015-8-3 10:49:53 File Name :C.cpp *************************************************/ #in…

题目传送门 /* 构造:首先先选好k个不同的值,从1到k,按要求把数字放好,其余的随便放.因为是绝对差值,从n开始一下一上, 这样保证不会超出边界并且以防其余的数相邻绝对值差>k */ /************************************************ Author :Running_Time Created Time :2015-8-2 9:20:01 File Name :B.cpp **************************************…

题目传送门 /* 构造:从大到小构造,每一次都把最后不是9的变为9,p - p MOD 10^k - 1,直到小于最小值. 另外,最多len-1次循环 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long ll; ; const int INF = 0x3f3f3f3f; int…

题目传送门 /* 构造:结构体排个序,写的有些啰嗦,主要想用用流,少些了判断条件WA好几次:( */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <map> #include <iostream> #include <string> using name…

A. Calculating Function 水题,分奇数偶数处理一下就好了 #include<stdio.h> #include<iostream> using namespace std; int main() { long long n;scanf("%lld",&n); ==) printf("%lld\n",(n-1LL)/2LL - n); else printf("%lld\n",n/2LL); }…

题目地址:http://codeforces.com/contest/486 A题.Calculating Function 奇偶性判断,简单推导公式. #include<cstdio> #include<iostream> using namespace std; int main() { long long n; cin>>n; ==) { cout<<(-)*((n-)/+)+n<<endl; } else cout<<((n-…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud A. Calculating Function 水题,判个奇偶即可 #include <iostream> #include <sstream> #include <ios> #include <iomanip> #include <functional> #include <algorithm> #include &…

B. OR in Matrix Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem/B Description Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if eith…

A. Calculating Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn Your task is to calculate f(n…

#include<iostream> #include<cstring> #include<cstdio> /* 题意:计算f(n) = -1 + 2 -3 +4.....+(-1)^n *n的值 思路:偶数和 - 奇数和(或者用等差数列计算化简得到结果) */ #include<algorithm> #define N 10000 using namespace std; int main(){ long long n; cin>>n; ==)…

整理上次写的题目: A: For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn Your task is to calculate f(n) for a given integer n. Input The single line contains the positive integer n (1 ≤ n ≤ 10^15). 题目简洁.可以看出规律...分下奇偶就可以了.…

E. LIS of Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem/E Description The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence.…

D. Valid Sets Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem/D Description As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consist…

A. Calculating Function Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem/A Description For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn Your task is to calculate f(n) for a…

题目链接:www.codeforces.com/problemset/problem/486/A题意:求表达式f(n)的值.(f(n)的表述见题目)C++代码: #include <iostream> using namespace std; long long f(long long n) { == ) ; else - n; } int main() { long long n; cin >> n; cout << f(n) << endl; ; } C…

Points on Plane Problem's Link Mean: 在二维坐标中给定n个点,求一条哈密顿通路. analyse: 一开始忽略了“无需保证路径最短”这个条件,一直在套最短哈密顿通路的模板,无限TLE. 简单的构造,首先对x坐标设一个阀值,分段输出,从下到上.再从上到下.在从下到上...直到所有点输出完为止. 当然也可横向扫描输出. Time complexity: O(N) Source code:  ;;;;)                  ;}…

题意:让你构造一个只包含小写字母的可重集,每次可以取两个元素,将它们合并,合并的代价是这两个元素各自的从‘a’到‘z’出现的次数之积的和. 给你K,你构造的可重集必须满足将所有元素合而为一以后,所消耗的最小代价恰好为K. 考虑只包含一种类字母的消耗代价,以a为例: a 0 aa 1 aaa 3 aaa 6 aaaa 10 aaaaa 15 ... ... 而且如果再其上任意叠加别的字母的话,是互不干涉的.于是可以贪心地从K中依次减去最大的一个上表中的数,输出那么多‘a’,然后下一次换成'b',如…

对那个树进行dfs,在动态维护那个当前的冰激凌集合的时候,显然某种冰激凌仅会进出集合各一次(因为在树上形成连通块). 于是显然可以对当前的冰激凌集合贪心染色.暴力去维护即可.具体实现看代码.map不必要. #include<cstdio> #include<set> #include<vector> #include<map> #include<algorithm> using namespace std; map<int,bool>…

贪心地一个一个尽可能往口袋里放,容易发现和顺序无关. #include<cstdio> #include<iostream> using namespace std; typedef long long ll; int n,m,a[100100]; ll ans; int main(){ // freopen("a.in","r",stdin); scanf("%d%d",&n,&m); for(int i…

A:SwapSort http://codeforces.com/problemset/problem/489/A 题目大意:将一个序列排序,可以交换任意两个数字,但要求交换的次数不超过n,输出任意一种方案即可 思路:真心想复杂了,排个序,然后按顺序把每个数交换到它应该在的地方即可 #include <stdio.h> #include <algorithm> #define maxn 40000 using namespace std; struct T{int x;int y;…