题意:给出很多对数字,看看每一对(a,b)能不能找到对应的(b,a). 放在贪心这其实有点像检索. 用stl做,map+pair. 记录每一对出现的次数,然后遍历看看对应的那一对出现的次数有没有和自己出现的此时一样即可. 代码: /* * Author: illuz <iilluzen@gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: uva10763.cpp * Lauguage: C/C++ * Create Date: 2013-…

题目连接:10763 Foreign Exchange 题目大意:给出交换学生的原先国家和所去的国家,交换成功的条件是如果A国给B国一个学生,对应的B国也必须给A国一个学生,否则就是交换失败. 解题思路: 给出数据  10 x  y 1 2 2 1 3 4 4 3 100 200 200 100 57 2 2 57 1 2 2 1 按照排序: xy y x 12 1 2 12 1 2 21 2 1 21 2 1 257 2 57   34 3 4  43 4 3 572 57 2 100 200…

Foreign Exchange Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance w…

  Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Description Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the…

Your non-profitorganization (iCORE - international Confederationof Revolver Enthusiasts) coordinates a very successfulforeign student exchange program. Over the last few years, demand hassky-rocketed and now you need assistance with your task. The pr…

题目:10763 - Foreign Exchange 题目大意:给出每一个同学想要的交换坐标 a, b 代表这位同学在位置a希望能和b位置的同学交换.要求每一位同学都能找到和他交换的交换生. 解题思路:把给定的原先给定的序列,交换前后位置后得到新的序列. 假设这个新的序列和原来的序列同样就符合要求.由于  (a.b) (b. a)若是成对出现.那么前后交换后还是(b. a)(a.b). 代码: #include <stdio.h> #include <string.h> #inc…

题目 题目     分析 没什么好说的,字符串拼接一下再放进map.其实可以直接开俩数组排序后对比一下,但是我还是想熟悉熟悉map用法. 呃400ms,有点慢.     代码 #include <bits/stdc++.h> using namespace std; int read(string a) { int ans=0,i=0; while(i<a.length()){ans=(ans<<3)+(ans<<1)+a[i]-'0';i++;} return…

 10763 Foreign ExchangeYour non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinatesa very successful foreign student exchange program. Over the last few years, demand hassky-rocketed and now you need assistan…

Foreign Exchange Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over th…

 Foreign Exchange Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance…

题意:有若干交换生.若干学校,有人希望从A校到B校,有的想从B到C.C到A等等等等.如果有人想从A到B也刚好有人想从B到A,那么可以交换(不允许一对多.多对一).看作后如果有人找不到人交换,那么整个交换计划失败. 代码:(Accepted, 50ms) //UVa10763 - Foreign Exchange #include<cstdio> #include<cstring> int N, a, b,all[1003][1003]; int main() { //freopen…

  Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.…

看到大神说location的值不会超过1000,所以这就简单很多了,用一个deg数组记录下来每个点的度,出度-1,入读+1这样. 最后判断每个点的度是否为0即可. 至于为什么会这样,据说是套数据套出来的,比如在代码里加一句if(a >= 1000) for(;;),get新技能! 如果按正常来做的话,我能想到的就是遍历map了. #include <cstdio> #include <cstring> ; int deg[maxn]; int main() { //freop…

10763:水题不解释直接贴代码. #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <math.h> #include <queue> #define eps 1e-9 typedef long long ll; using namespace std; int n; ]; int main() { int…

题目描述: 解题思路: 开一个数组,读入一次交换两个数,如果最后数组不变,即符合匹配 #include<iostream> #include<cstdio> #include<string> #include<algorithm> #define MAXN 500005 using namespace std; int arr[MAXN]; void swap(int a,int b){ int t=arr[a]; arr[a] = arr[b]; arr[…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 各组数据之间有空行! 且最后一行后面没有空行! 然后就是用set来模拟就好. 删除的时候,不着急删除. 因为并不用时刻输出集合大小.所以只要遇到了把它删掉就Ok. 把相同的合并那里.我直接暴力合并了. 因为 150 30 100 30 不能看出一个整体的250 30的.. 要分步输出Trade信息的. 然后在合并的时候也要注意看里面有没有已经删除了的. 已经删除了的就直接跳过.. [代码] #include <bits/stdc…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 如果x>y 则num[(x,y)]--; 否则num[(x,y)]++; 看看每一个二元组的num值是不是都为0就好. [代码] #include <bits/stdc++.h> using namespace std; int n; map <pair<int, int>, int> mmap; int main() { //freopen("F:\\rush.txt", &…

题目链接: https://www.luogu.org/problemnew/show/UVA10763 题目分析: 本题我首先想到的做法是把每一个数都map一下,然后互相判断,例如a,b两人准备交换,那么m[a]=b,m[b]=a,最后再判断如果m[m[a]]=a就行,但是因为一个学生在双方都同意的情况下可以先后与多个学生交换,所以我的做法是: 先把每对学生都按照先小后大的顺序排好 if(a[i].x>a[i].y)swap(a[i].x,a[i].y); 然后把所以学生交换对都进行排序 so…

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例题5--9 数据库 Database UVa 1592 #include<iostream> #include<stdio.h> #include<string.h> #include<cmath> #include<string> #include<queue> #include<stack> #include<vector> #include<map> #include<set>…

MENU: Transactions > Financial > Revalue Open Currency Balances 使用频率: - 每个月月底,结账前, 手工操作. - 或者在Period Close Checklist会有一步需要操作:Revalue Open Foreign Currency Balances Note: - 页面Month End Currency Revaluation,每个Subsidiary都要做,在列表下面系统会列出所有等待有可能存在交易外币金额的科目…

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英语新闻常用词汇与短语 经济篇 accumulated deficit 累计赤字 active trade balance 贸易顺差 adverse trade balance 贸易逆差 aid 援助 allocation of funds 资金分配 allotment 拨款 allowance/grant/subsidy 补贴,补助金,津贴 amortization 摊销,摊还,分期偿付 annuity 年金 article 物品,商品 assigned 过户 autarchy 闭关自守 av…

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1, Index An index is a set of data pointers stored on disk associated with a single table. The main advantage is they greatly speed up select, update, and delete statements, as the query optimizer only has to search the index rather than the entire t…

UNIX or Linux operating system has become default Server operating system and for whichever programming job you give interview you find some UNIX command interview questions there. These UNIX command interview questions are mostly asked during Java d…

my.oschina.net/fcboys/blog/2695 FXS (Foreign Exchange Station) FXS is an interface which drives a telephone. FXS interfaces get phones plugged into them, deliver battery, and provide ringing. The FXS interface initiates and sends a ringing voltage to…

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A curated list of awesome Python frameworks, libraries, software and resources. Inspired by awesome-php. Admin Panels Libraries for administrative interfaces. Ajenti - The admin panel your servers deserve. django-suit - Alternative Django Admin-Inter…