LightOj 1148 Basic Math

1148 - Mad Counting

PDF (English) Statistics Forum

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





PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE

SPECIAL THANKS: JANE ALAM JAN

思路：

把每一个人说的数和说这个数的人数分别存在了两个数组中。然后用每一个数除以这个数被说的次数向上取整累计加和就可以。

<span style="color:#3366ff;">/***********************************
    author   : Grant Yuan
    time     : 2014/8/21 10:22
    algorithm: Basic Math
    source   : LightOj 1148
************************************/
#include<bits/stdc++.h>

using namespace std;
int a[57];
int f[57];
int main()
{
    int t,n,ans,sum;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        memset(a,0,sizeof(a));
        memset(f,0,sizeof(f));
        scanf("%d",&n);
        ans=0;sum=0;
        for(int j=0;j<n;j++)
        {
            int m;
            scanf("%d",&m);
            if(!binary_search(a,a+sum,m+1))
            {

                a[sum]=m+1;
                f[sum++]++;
            }
            else
            {
                for(int k=0;k<sum;k++)
                {
                    if(a[k]==m+1) f[k]++;
                }
            }
        }
       for(int j=0;j<sum;j++)
       {
           ans+=((f[j]+a[j]-1)/a[j])*a[j];
       }
        printf("Case %d: %d\n",i,ans);
    }
    return 0;
}
</span>

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