1148 - Mad Counting(数学)
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 |
题解:没理解清题意,看见世界杯,就想着中国队和巴西队,然而并没有这么少的队;
其实吧,这个题就是说,问卷调查,问除了你还有几个人和你支持同一个队;本渣的思路是:统计x的相同的个数为y,因为,两个人支持同一个队,人数必然相同,那么让这y个x组最小的队,一个x再要x个人可以组一个队,那么找y里面有几个x+1即可,那么代码救出来了;
代码:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=1e6+;
int pre[MAXN];
int a[MAXN];
int main(){
int T,N,temp,flot=;
scanf("%d",&T);
while(T--){
memset(pre,,sizeof(pre));
scanf("%d",&N);
int pos=;
for(int i=;i<N;i++){
scanf("%d",&temp);
if(!pre[temp])
a[pos++]=temp;
pre[temp]++;
}
int ans=,x,y;
for(int i=;i<pos;i++){
x=a[i];y=pre[a[i]];
ans+=y/(x+)*(x+);
if(y%(x+)!=)ans+=x+;
}
printf("Case %d: %d\n",++flot,ans);
}
return ;
}
再次出了这个题,代码;
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define P_ printf(" ")
#define T_T while(T--)
typedef long long LL;
const int INF=0x3f3f3f3f;
const int MAXN=1e6+100;
int m[MAXN];
int main(){
int T,kase=0,N;
SI(T);
T_T{
SI(N);
int x;
for(int i=0;i<N;i++){
SI(x);
m[i]=x+1;
}
sort(m,m+N);
int temp=m[0],flot=1,sum=m[0];
for(int i=1;i<N;i++){
if(m[i]!=m[i-1]){
flot=1;temp=m[i];
sum+=temp;
continue;
}
flot++;
if(flot>temp)sum+=temp,flot=1; }
printf("Case %d: %d\n",++kase,sum);
}
return 0;
}
1148 - Mad Counting(数学)的更多相关文章
- 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 ...
- light oj 1148 - Mad Counting
1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB M ...
- LightOJ - 1148 - Mad Counting
先上题目: 1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 3 ...
- Codeforces 911D. Inversion Counting (数学、思维)
题目链接:Inversion Counting 题意: 定义数列{ai|i=1,2,...,n}的逆序对如下:对于所有的1≤j<i≤n,若ai<aj,则<i,j>为一个逆序对. ...
- 17997 Simple Counting 数学
17997 Simple Counting 时间限制:2000MS 内存限制:65535K提交次数:0 通过次数:0 题型: 编程题 语言: 不限定 Description Ly is craz ...
- UVA 11401 - Triangle CountingTriangle Counting 数学
You are given n rods of length 1,2, . . . , n. You have to pick any 3 of them and build a triangle. ...
- LightOj 1148 Basic Math
1148 - Mad Counting PDF (English) Statistics Forum Time Limit: 0.5 second(s) Memory Limit: 32 MB Mob ...
- acdream.A Very Easy Triangle Counting Game(数学推导)
A - A Very Easy Triangle Counting Game Time Limit:1000MS Memory Limit:64000KB 64bit IO Forma ...
- UVALive 5058 Counting BST 数学
B - Counting BST Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit S ...
随机推荐
- rsa 密钥和公钥的生成
openssl genrsa -out prikey.pem openssl rsa -in prikey.pem -pubout -out pubkey.pem
- JSTL入门
在页面最上方引入 -------------------- if语句 8}"> b的值大于8 --------------------- foreach语句 i的值是:${i}
- hdu3033I love sneakers! (分组背包,错了很多次)
Problem Description After months of hard working, Iserlohn finally wins awesome amount of scholarshi ...
- ThinkPHP第二十二天(表单令牌、相对路径、扩展配置载入、$Think获取系统变量、$_SERVER('HTTP_REFERER')前页地址)
1.表单令牌开启配置 'TOKEN_ON'=>true 2.相对路径:在thinkphp中,存在单入口index.php,所以程序中的根目录都是以index.php所在的文件夹为根目录,故用./ ...
- codeforces 463C Gargari and Bishops
题目链接 这个题, 最主要的应该是找到对角线上的格子的关系. “ \" 这种对角线, 关系是x-y+n相等, ” / “ 这种, 关系是x+y相等.知道每个格子的两种对角线的值, 那么这个格 ...
- 【LeetCode题意分析&解答】41. First Missing Positive
Given an unsorted integer array, find the first missing positive integer. For example,Given [1,2,0] ...
- (Problem 40)Champernowne's constant
An irrational decimal fraction is created by concatenating the positive integers: 0.1234567891011213 ...
- Make Yahoo! Web Service REST Calls With C#
原文 http://developer.yahoo.com/dotnet/howto-rest_cs.html The .NET Framework provides classes for perf ...
- axis1客户端调用webservice的通用代码
1.axis1 作为web service 客户端时,调用web service 服务端的通用代码 String url = "http://www.webxml.com.cn/webser ...
- 有道翻译API
轻奢侈品_百度百科 轻奢侈品 有道翻译API 有道翻译API申请成功 API key:72763558 keyfrom:lexus-studio