Greedy

http://codeforces.com/contest/12/problem/C 题意：有n个价格价格，m个要买的东西（可能有相同的种类，设为k种），把n个标签中拿出k个给个贴上。。。问最大价钱和最少价钱分别是多少。 思路：贪心。不过要按照map的value排序。。然后发现其实不用排序。。因为map的key值其实不影响。

http://codeforces.com/contest/526/problem/B 题意：有一棵完全二叉树。每条边上有一定数量的路灯。问最少需要添加多少个路灯。使得根节点道叶子节点的每一条路径上的路灯数量一样。 思路：同叶子节点网上更新即可。

http://codeforces.com/problemset/problem/596/B 题意：初始序列全为0，问经过多少次变换，能变成序列b。一次变换是指，选定一个i，从i一直到最后每个元素都增加1，或者每个元素都减少1. 思路：很容易发现。后面的变换补影响前面的变换。每一个数字可以唯一由之前的增加和减少次数决定。所以我们用两个变量，记录之前做的增加和减少变换的次数。然后扫一遍即可。

http://codeforces.com/problemset/problem/600/C 题意：给定一个字符串。要求用最少的变换得到一个回文串。且在变换次数相同时要字典序最小的。输出变换后的字符串。 思路：对不能构成会文串有影响的是出现奇数次的字母。所以我们先统计每个字母出现的次数。然后按照出现奇数次的字母的个数分奇偶分别搞。偶数的话直接把后面一半变成前面一半。奇数的话，也是这样。输出的时候按照字母从a扫到z，如果有就输出一半。然后再倒着扫一遍。 输出另一半。这样可以保证是字典序最小。需要注意的是奇数的时候的输出情况。不要忘记中间那个字母。

http://codeforces.com/problemset/problem/158/B 题意：n组人，每组有si个（1<=si<=4），每辆车能装4个人。问最少需要多少辆车装下所有人并且保证同一组的人在一辆车里。 思路：统计人数分别为1,2,3,4的人数。对于4的直接加到答案。贪心的思路是：优先用人数少的去填人数多的。

题意是说。给n个balls,k个箱子。保证（n<=2*k） 一个箱子中中最多放两个balls，size为两个balls的size之和。 所有的箱子的size都要一样。 问size最小是多少。

贪心．．尽量把一样的材料放在一起．．． 然后写蠢了．．妈蛋．．． 详情见代码

http://codeforces.com/contest/558/problem/C 题目大意是说，给定N个数，可以对任意数进行任意次两种操作，×2，和/2（整除）

http://codeforces.com/problemset/problem/479/D 题意是说有一把尺子，本身有一些刻度，然后需要测量x和y，问最少需要添加多少个刻度，如果需要，这些刻度分别添加在什么位置。

http://codeforces.com/problemset/problem/525/B 1题意是说一个字符串，进行m次颠倒变换（从a[i]位置到a[l-i+1]位置），问得到的字符串。容易发现，对于越在里边（对称，也就是越靠近中间位置）的字符，调换的次数越多。我们可以把a[i]从小到大排序。然后经过分析发现，把两个相邻的a[i]分为一组，做处理，如果m为奇数，最后还剩下a[m]没有被分组，要单独处理a[m]细节上要注意st数组是从st[0]开始的...好吧的确不方便，适牛也说我了。。数组下标以后还是从0开始吧。。。主要是受高中OI用的pascal的影响。。。那个数组下标随便啊。代码： 2 3 4 5 6 /* *********************************************** 7 Author :111qqz 8 Created Time :2016年02月22日 星期一 23时39分51秒 9 File Name :code/cf/problem/525B.cpp 10 ************************************************ */ 11 12 #include <iostream> 13 #include <algorithm> 14 #include <cstring> 15 #include <cmath> 16 #include <cstdio> 17 18 using namespace std; 19 20 int m,k,len; 21 const int N=2E5+7; 22 int a[N]; 23 char st[N]; 24 25 int main() 26 { 27 cin>>st; 28 scanf("%d",&m); 29 for ( int i = 1 ; i <= m ; i++ ) 30 scanf("%d",&a[i]); 31 sort(a+1,a+m+1); 32 k = 1; 33 len = strlen(st); 34 while (k<=m) 35 { 36 for ( int j = a[k] ; j <= a[k+1]-1 ; j++) 37 swap(st[j-1],st[len-j]); 38 k = k + 2; 39 } 40 if ( m %2==1 ) 41 for ( int i = a[m]; i <= len/2 ; i++ ) 42 swap(st[i-1],st[len-i]); 43 cout<<st<<endl; 44 return 0; 45 }

C - C **Time Limit:**1000MS **Memory Limit:**262144KB 64bit IO Format:%I64d & %I64u Submit Status Description Permutation_p_ is an ordered set of integers _p_1, p_2, …, p__n, consisting of n distinct positive integers not larger than n. We’ll denote as_n the length of permutation _p_1, _p_2, …, p__n.

http://codeforces.com/problemset/problem/479/C 1/************************************************ 2Author :111qqz 3Created Time :2016年02月22日 星期一 23时31分10秒 4File Name :code/cf/problem/479C.cpp 5************************************************ */ 6 7#include <iostream> 8#include <algorithm> 9#include <cstring> 10#include <cstdio> 11 12#include <cmath> 13 14using namespace std; 15int n,ans; 16const int N=1E4+5; 17int a[N],b[N]; 18 19struct Q 20{int a,b; 21}q[N]; 22 23bool cmp(Q x, Q y) 24{ 25 if ( x.a<y.a) return true; 26 if ( x.a==y.a &&x.b<y.b ) return true; 27 return false; 28} 29 30int main() 31{ 32 scanf("%d",&n); 33 for ( int i = 1 ; i <= n ; i++ ) 34 scanf("%d %d",&q[i].a,&q[i].b); 35 sort(q+1,q+n+1,cmp); 36 37 ans=q[1].b; 38 for ( int i = 2 ; i <= n; i++ ) 39 { 40 if ( q[i].b>=ans ) 41 ans = q[i].b; 42 else ans = q[i].a; 43 } 44 printf("%d\n",ans); 45 return 0; 46}

ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 988 Accepted Submission(s): 500 Problem Description Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called “Memset137”. It was on Codefires(CF), an online competitive programming site, that ZCC knew Memset137, and immediately became his fan. But why? Because Memset137 can solve all problem in rounds, without unsuccessful submissions; his estimation of time to solve certain problem is so accurate, that he can surely get an Accepted the second he has predicted. He soon became IGM, the best title of Codefires. Besides, he is famous for his coding speed and the achievement in the field of Data Structures. After become IGM, Memset137 has a new goal: He wants his score in CF rounds to be as large as possible. What is score? In Codefires, every problem has 2 attributes, let’s call them Ki and Bi(Ki, Bi＞0). if Memset137 solves the problem at Ti-th second, he gained Bi-KiTi score. It’s guaranteed Bi-KiTi is always positive during the round time. Now that Memset137 can solve every problem, in this problem, Bi is of no concern. Please write a program to calculate the minimal score he will lose.(that is, the sum of Ki*Ti).

Wooden Sticks Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 19008 Accepted: 8012 Description There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: (a) The setup time for the first wooden stick is 1 minute. (b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l’ and weight w’ if l <= l’ and w <= w’. Otherwise, it will need 1 minute for setup. You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) . Input

简单贪心。 因为填的字母没有次数限制，所以最优策略很容易想到，就是在最后面填最大的。 不用实际去填，算出ans就可以。

简单贪心…. 需要注意的是数据是非负，所以有0的情况要考虑周全，基本都要特殊处理。 多WA了三次，不知道为什么交C++可以过，交G++就不行。