light oj 1148 - Mad Counting

Mad Counting

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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

 

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PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE

SPECIAL THANKS: JANE ALAM JAN

题意：民意调查，每一名公民都有盟友，问最少人数。

排序，如果被调查的人有相同盟友人数（n）的个数（cnt）大于这个数+1，即：（cnt > n+1）, 该共同的盟友的数量已经超过了盟友容量限度，只能新开来装盟友。

例如  ：a 说b是他的盟友！

      b说a是他的盟友！   此时容量为2 ！最多只能有两个人共同在该盟友对中！

代码：  

     

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int  main(){int  T;long long  a[1000];scanf("%d",&T);int S=T;while(T--){int  n,i;scanf("%d",&n);for(i=0;i<n;i++){scanf("%lld",&a[i]);}sort(a,a+n);long long   all=0,num=0;for(i=0;i<n;i++){       if(i==0||a[i]==a[i-1])       num++;//记录该数出现的次数        else       {       int  t=(num+a[i-1])/(a[i-1]+1);//num+a[i-1] 是看一共有多少 出现次数，并向上取整，a[i-1]+1该盟友集合的容量        all+=t*(a[i-1]+1);//取整后进行就按        num=1;   }}all+=(num+a[i-1])/(a[i-1]+1)*(a[i-1]+1);printf("Case %d: %lld\n",S-T,all);}return  0; }