light oj 1148 - Mad Counting
| Time Limit: 0.5 second(s) | Memory Limit: 32 MB |
Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this problem will be counting people one by one. But as we all know Mob is a bit lazy, so he is finding some other approach so that the time will be minimized. Suddenly he found a poll result of that town where N people were asked "How many people in this town other than yourself support the same team as you in the FIFA world CUP 2010?" Now Mob wants to know if he can find the minimum possible population of the town from this statistics. Note that no people were asked the question more than once.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with an integer N (1 ≤ N ≤ 50). The next line will contain N integers denoting the replies (0 to 106) of the people.
Output
For each case, print the case number and the minimum possible population of the town.
Sample Input |
Output for Sample Input |
|
2 4 1 1 2 2 1 0 |
Case 1: 5 Case 2: 1 |
题意:现在要求一个村庄的总人数,向n个人询问在他们的村庄有多少人和他自己喜欢同一个球队根据数据估计村庄的最少人数
题解:输入数据对数据从小到大排序,如果有连续相同的数a则将这些连续相同的看做是喜欢的同一个球队的人,(相同的个数最多是a+1 如果超过这个则按另一支队算)
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1000
#define MAXM 1001000
#define INF 0x7fffff
#define LL long long
using namespace std;
int s[MAXM];
int main()
{
int n,m,j,i,t,k;
int sum;
//LL sum;
scanf("%d",&t);
k=1;
while(t--)
{
sum=0;
int ans;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&s[i]);
sort(s,s+n);
for(i=0;i<n;i++)
{
int num=0;
ans=s[i];
num=s[i];
while(s[i+1]==ans&&num)
{
i=i+1;
num--;
}
sum=sum+ans+1;
}
printf("Case %d: ",k++);
printf("%d\n",sum);
}
return 0;
}
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1000
#define MAXM 1001000
#define INF 0x7fffff
#define LL long long
using namespace std;
int s[MAXM];
int main()
{
int n,m,j,i,t,k;
int sum;
//LL sum;
scanf("%d",&t);
k=1;
while(t--)
{
sum=0;
int ans;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&s[i]);
sort(s,s+n);
for(i=0;i<n;i++)
{
int num=0;
ans=s[i];
num=s[i];
while(s[i+1]==ans&&num)
{
i=i+1;
num--;
}
sum=sum+ans+1;
}
printf("Case %d: ",k++);
printf("%d\n",sum);
}
return 0;
}
light oj 1148 - Mad Counting的更多相关文章
- 1148 - Mad Counting(数学)
1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB M ...
- lightoj 1148 Mad Counting(数学水题)
lightoj 1148 Mad Counting 链接:http://lightoj.com/volume_showproblem.php?problem=1148 题意:民意调查,每一名公民都有盟 ...
- LightOJ - 1148 - Mad Counting
先上题目: 1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 3 ...
- Light OJ - 1058 Parallelogram Counting(判定平行四边形)
Description There are n distinct points in the plane, given by their integer coordinates. Find the n ...
- Light OJ 1148
题意: 给你N 个人, 每个人说出有多少人和他一队, 不包括他自己, 输出总人数最少值 思路: 排个序, 按照给的数目把人分为一组,就可以得出最少人数 #include<bits/stdc++. ...
- Light OJ 1114 Easily Readable 字典树
题目来源:Light OJ 1114 Easily Readable 题意:求一个句子有多少种组成方案 仅仅要满足每一个单词的首尾字符一样 中间顺序能够变化 思路:每一个单词除了首尾 中间的字符排序 ...
- Light OJ 1429 Assassin`s Creed (II) BFS+缩点+最小路径覆盖
题目来源:Light OJ 1429 Assassin`s Creed (II) 题意:最少几个人走全然图 能够反复走 有向图 思路:假设是DAG图而且每一个点不能反复走 那么就是裸的最小路径覆盖 如 ...
- Light OJ 1406 Assassin`s Creed 减少国家DP+支撑点甚至通缩+最小路径覆盖
标题来源:problem=1406">Light OJ 1406 Assassin`s Creed 意甲冠军:向图 派出最少的人经过全部的城市 而且每一个人不能走别人走过的地方 思路: ...
- Light OJ 1316 A Wedding Party 最短路+状态压缩DP
题目来源:Light OJ 1316 1316 - A Wedding Party 题意:和HDU 4284 差点儿相同 有一些商店 从起点到终点在走过尽量多商店的情况下求最短路 思路:首先预处理每两 ...
随机推荐
- BZOJ 1688: [Usaco2005 Open]Disease Manangement 疾病管理
Description Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the fa ...
- sqlserver删除重复数据
select identity(int,1,1) as autoID, * into #Tmp from [QYTS_QiYeCecdb]select min(autoID) as autoID in ...
- [TSP+floyd]POJ3311 Hie with the Pie
题意: 给i到j花费的地图 1到n编号 一个人要从1遍历n个城市后回到1 求最小的花费(可以重复走) 分析 http://www.cnblogs.com/Empress/p/4039240.htm ...
- [codility]Prefix-set
这题很简单,一开始用了set.但后来一想这样其实是n*logn的,而且没有利用所有的数都在0..N-1之间.那么可以直接用vector当hashset. // you can also use inc ...
- UC何小鹏:移动互联网创业需警惕五大“不靠谱
http://tech.qq.com/a/20140121/012443.htm 腾讯科技 启言 1月21日报道 移动互联网创业很容易犯错误,一不小心就陷入“坑”中.UC也是如此.近日,UC创始人何小 ...
- 微小说《tree》
自己一直是一个矛盾纠结的人,计算机专业,却喜欢画画,偶尔会写些短文,寓言,或者酸酸的情诗.. 废话不多说,各位看官有兴趣便随意看看,若有些于共鸣,便共勉之. 正文 两株幼苗从土壤中破土而出,一颗天生强 ...
- windows 上rails3.2 + ruby1.9环境搭建
题外话:本文是通过参考网友资料,亲自尝试过后写的,有不对之处,还请网友指正! 1.搭建环境 准备ruby1.9.3 下载地址: 下载地址:http://rubyforge.org/frs/?group ...
- MySQL 线上配置文件
[client] port = 3306 socket = /tmp/mysql.sock default-character-set = utf8 [mysq ...
- 海量数据的二度人脉挖掘算法(Hadoop 实现)
最近做了一个项目,要求找出二度人脉的一些关系,就好似新浪微博的“你可能感兴趣的人” 中,间接关注推荐:简单描述:即你关注的人中有N个人同时都关注了 XXX . 在程序的实现上,其实我们要找的是:若 U ...
- MapReduce架构设计
MapReduce采用Master/Slave的架构,其架构图如下: 它主要有以下4个部分组成: 1)Client 2)JobTracker JobTracke负责资源监控和作业调度.JobTrack ...