[简要题意]:这个主题是很短的叙述性说明.挺easy. 不重复. [分析]:只需要加一个判断这个数是否可以是一个数组,这个数组的范围. // 3388K 0Ms #include<iostream> using namespace std; #define Max 500001 int a[Max]; bool b[10000000] = {false}; // b的数据范围是能够试出来的- void init() { a[0] = 0; b[0] = true; for(int m = 1;…

開始还以为暴力做不出来,须要找规律,找了半天找不出来.原来直接暴力.. 代码例如以下: #include<stdio.h> int a[1000050]; int b[100000000]={0}; int main() { int i,k; a[0]=0; for(i=1;i<=500000;i++) { a[i]=a[i-1]-i; if(a[i-1]-i>0&&!b[a[i-1]-i]) a[i]=a[i-1]-i; else a[i]=a[i-1]+i; b…

Recaman's Sequence Time Limit: 3000ms Memory Limit: 60000KB This problem will be judged on PKU. Original ID: 208164-bit integer IO format: %lld      Java class name: Main   The Recaman's sequence is defined by a0 = 0 ; for m > 0, am = am−1 − m if the…

                                                                                                      Recaman's Sequence Time Limit: 3000MS   Memory Limit: 60000K Total Submissions: 22363   Accepted: 9605 Description The Recaman's sequence is defined…

Recaman's Sequence Time Limit: 3000MS   Memory Limit: 60000K Total Submissions: 22566   Accepted: 9697 Description The Recaman's sequence is defined by a0 = 0 ; for m > 0, am = am−1 − m if the rsulting am is positive and not already in the sequence,…

UVa 1584 题目大意:给定一个含有n个字母的环状字符串,可从任意位置开始按顺时针读取n个字母,输出其中字典序最小的结果 解题思路:先利用模运算实现一个判定给定一个环状的串以及两个首字母位置,比较二者字典序大小的函数, 然后再用一层循环,进行n次比较,保存最小的字典序的串的首字母位置,再利用模运算输出即可 /* UVa 1584 Circular Sequence --- 水题 */ #include <cstdio> #include <cstring> //字符串s为环状,…

A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/622/problem/A Description Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is wr…

A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/675/problem/A Description Vasya likes everything infinite. Now he is studying the properties of a sequence s, such that its first element is equal to a (s1 = a), and the difference between…

LINK 题意:给出一个简单多边形,按极角序输出其坐标. 思路:水题.对任意两点求叉积正负判断相对位置,为0则按长度排序 /** @Date : 2017-07-13 16:46:17 * @FileName: POJ 2007 凸包极角序.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <…

题目大意:给出一些海报和贴在墙上的区间.问这些海报依照顺序贴完之后,最后能后看到多少种海报. 思路:区间的范围太大,然而最多仅仅会有10000张海报,所以要离散化. 之后用线段树随便搞搞就能过. 关键是离散化的方法,这个题我时隔半年才A掉,之前一直就TTT,我还以为是线段树写挂了. 当我觉得我自己的水平这样的水线段树已经基本写不挂的时候又写了这个题,竟然还是T. 后来我对照别人的代码,才发现是我的离散化写渣了. 以下附AC代码(79ms),这个离散化写的比較优雅.时间也非常快,以后就这么写了.…