题目链接:C. Soldier and Cards Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between them in some manner, it's possible that they have different numb…

题目链接 题意 两个人玩扑克,共n张牌,第一个人k1张,第二个人k2张 给定输入的牌的顺序就是出牌的顺序 每次分别比较两个人牌的第一张,牌上面数字大的赢,把这两张牌给赢的人,并且大的牌放在这个人的牌最下面,另外一张放在上面牌的上面,其他牌在放在这两张牌的上面. 求要pk多少次结束游戏,并记录赢得是哪个人 若出现死循环的情况输出 –1   这里可以根据栈或队列 java的程序是根据栈的,比较时候取出栈顶,加入新的两个 数的时候,要先出栈,在入栈,有点麻烦 Python程序是根据队列,在头取出进行比…

CodeForces 546C Soldier and Cards Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u   Description Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are d…

Soldier and Cards 老样子,直接上国语吧  Descriptions: 两个人打牌,从自己的手牌中抽出最上面的一张比较大小,大的一方可以拿对方的手牌以及自己打掉的手牌重新作为自己的牌,放在自己手牌的最下方,而且对方输掉的那张手牌需要放在上面,自己赢的手牌放在下面. Input 第一行的数n代表一共有几张牌 第二行第一个数x代表第一个人有x张牌 第三行第一个数y代表第二个人有y张牌 Output 第一个数代表进行了几轮,第二个数代表谁赢 Examples Input 42 1 32…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divid…

题目链接:http://codeforces.com/problemset/problem/546/C 题解: 用两个队列模拟过程就可以了. 特殊的地方是:1.如果等大,那么两张牌都丢弃 : 2.如果操作了很多次仍不能分出胜负,则认为平手.(至于多少次,我也不知道,只能写大一点碰运气,但要防止超时) 代码如下: #include<iostream>//C - Soldier and Cards #include<cstdio> #include<cstring> #in…

C. Soldier and Cards Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/546/problem/C Description Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are diff…

Soldier and Cards 题目: Description Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between them in some manner, it's possible that they have differ…

题目传送门 /* 题意:两堆牌,每次拿出上面的牌做比较,大的一方收走两张牌,直到一方没有牌 queue容器:模拟上述过程,当次数达到最大值时判断为-1 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <stack> #include <cmath> #inc…

Soldier and Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are…

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Description Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between…

题意 (Codeforces 546C) 按照指定的规则打牌,问谁胜或无穷尽. 分析 又是一条模拟,用set+queue(这里手写了)处理即可.注意到两种局势"1 234"和"123 4"的差别,所以用set处理的时候需要在两方手牌中间加上相关的分割符号以示区分. 代码 #include <bits/stdc++.h> #define MP make_pair #define PB push_back #define fi first #define s…

Codeforces 731 F. Video Cards 题目大意:给一组数,从中选一个数作lead,要求其他所有数减少为其倍数,再求和.问所求和的最大值. 思路:统计每个数字出现的个数,再做前缀和,用于之后快速求和.将原数组排序后去重,枚举每一个数做lead的情况下,其余数减少后再求和的结果.不断维护最大值即可. PS:这里用到了一个方便的函数unique()来去重,使用前需要先将数组排序,参数为数组的首地址和尾后地址,返回新的尾后地址.(该函数没有将重复的元素删除,只是放到了尾地址后面)可…

B. Ancient Berland Hieroglyphs 题目连接: http://codeforces.com/problemset/problem/164/B Descriptionww.co Polycarpus enjoys studying Berland hieroglyphs. Once Polycarp got hold of two ancient Berland pictures, on each of which was drawn a circle of hierog…

Problem G. Persistent QueueTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88258#problem/G Description Persistent data structures are designed to allow access and modication of any version of data st…

George and Cards 我们找到每个要被删的数字左边和右边第一个比它小的没被删的数字的位置.然后从小到大枚举要被删的数, 求答案. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pa…

The only printer in the computer science students' union is experiencing an extremely heavy workload. Sometimes there are a hundred jobs in the printer queue and you may have to wait for hours to get a single page of output. Because some jobs are mor…

题目链接:http://codeforces.com/problemset/problem/140/B 题目意思:给出 Alexander 和他的 n 个朋友的 preference lists:数字 1-n 的排列.现在Alexander 需要遵循两条rules向他的 朋友发送贺卡. 1.他不会send 和 该朋友给他的一模一样的贺卡. 2.对于当前他所拥有的贺卡,他只会选择他自己最喜欢的卡给朋友. 同一张卡可以使用多次,而且尽量使得他的朋友满意,也就是尽量满足该朋友的 preference…

Codeforces 题面传送门 & 洛谷题面传送门 首先先讲一发我的 \(n^2q\log n\) 的做法,虽然没有付诸实现并且我也深知它常数巨大过不去,但是我还是决定讲一讲(大雾 考虑设 \(f(i,j)\) 表示以 \(i,j\) 为左上角,满足其中不同颜色个数 \(\le q\) 的子正方形中大小最大的那个的边长,那么显然求出 \(f(i,j)\) 后一遍后缀和即可求出答案.于是问题转化为如何求 \(f(i,j)\).注意到一个显然的结论:\(f(i,j)\ge f(i,j-1)-1\)…

A. Team Olympiad time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The School №0 of the capital of Berland has n children studying in it. All the children in this school are gifted: some of t…

题目大意:给你一堆n张牌(数字可以相同),你只能从上面取牌,如果是当前牌堆里面最小的值则拿走, 否则放到底部,问你一共要操作多少次. 思路:讲不清楚,具体看代码.. #include<bits/stdc++.h> #define pb push_back #define ll long long using namespace std; ; ll n,mxi[N],a[N];//a[i]保存原始数据,mxi[i]保存大小为i的牌最下面一张的编号 vector<ll> p[N];//…

C. Bracket Subsequence ... 代码: 1 //C 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 #include<bitset> 7 #include<cassert> 8 #include<cctype> 9 #include<cmath> 10 #include&…

C. Page Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output «Bersoft» company is working on a new version of its most popular text editor - Bord 2010. Bord, like many other text editor…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately…

题目描述: Soldier and Number Game time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Two soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to t…

思路 是一道水题,可以用队列+模拟来写,注意不要拿完队列中的元素! 代码 #include<iostream> #include<cstdio> #include<queue> using namespace std; +; int main() { int n; cin>>n; queue<int>a;//队列 queue<int>b; int x,y; cin>>x; ;i<x;i++) { int x; cin…

题目链接:http://codeforces.com/problemset/problem/1172/A 题意:一共有2*n张牌,n张0,n张1到n.现在随机的n张(有0有数字)在手上,另n张再牌堆中,现在已知手上的牌和牌堆的牌,可以进行多次以下操作:将手中任意一张牌放入牌堆底,将牌堆顶的一张牌放入手中.问最少多少次后可使牌堆顶到牌堆底的n张牌分别为1,2,3...n. 思路:模拟,判断现有牌堆底能不能继续往下接,例如00123456.不能的话还是等1取出来再往下放. AC代码: #includ…

两个人玩牌,首先两个人都拿出自己手牌的最上面的进行拼点,两张拼点牌将都给拼点赢得人,这两张牌放入手牌的顺序是:先放对方的牌再放自己的.若最后有一个人没有手牌了,那么他就输了,求输出拼点的次数和赢得人的编号,如果一直无法结束比赛,则输出-1. 用队列模拟即可. #include <bits/stdc++.h> using namespace std; #define int long long #define ll long long #define reset(x) memset(x,0,si…

一个人只要存在债务关系,那么他的债务可以和这整个债务关系网中任何人连边,和他当初借出或欠下的人没有关系.只需要记录他的债务值即可. #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; vector<array<> >v; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); lon…

AC代码: #include<bits/stdc++.h> #define ll long long #define endl '\n' #define mem(a,b) memset(a,b,sizeof(a)) #define IO ios::sync_with_stdio(false);cin.tie(0); using namespace std; const int INF=0x3f3f3f3f; const ll inf=0x3f3f3f3f3f3f3f3f; ; ; struct…