LightOJ 1148 - Mad Counting【水题+规律】

1148 - 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

 

这个题不难，然而题意理解了很久.........

本来想着用数组来保存，然后发现数据太大，不太好，又想到用map，但是自己用的不太熟练，然后只能排序进行统计了....由于数据类型的转化没注意，一直出错，最后才解决......

一共有多少人，可以这样算：

如果一共有n个人说自己队有m个人，那么至少要有x=ceil（n/m）个队伍，那么一共要有x*m 个人，就这样把所有情况累加起来就得到了总人数.....

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
int t;
//freopen("shuju.txt","r",stdin);
scanf("%d",&t);
for(int i=1;i<=t;++i)
{
int n,x[105];
scanf("%d",&n);
for(int j=0;j<n;++j)
{
int tp;
scanf("%d",&tp);
x[j]=tp+1;//加上那个人
}
sort(x,x+n);//把相同的集合到一起
int sum=0,tp=0;
for(int j=0;j<n;++j)
{
int cnt=0;
while(j<n&&x[j]==x[tp])
{
++cnt;++j;
}
tp=j--;
//记录不同的首位置,下一次从第一个不同的开始统计
if(cnt<=x[j])//
{
sum+=x[j];
}
else
{
sum+=ceil(cnt*1.0/x[j])*x[j];//最少人数
}
}
printf("Case %d: %d\n",i,sum);
}
return 0;
}