lightOJ 1148 - Mad Counting 【水题】

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

PROBLEM SETTER: MUHAMMAD RIFAYAT SAMEE
SPECIAL THANKS: JANE ALAM JAN

思路：

首先将所有的数进行排序，然后将第一个数夹到手sum上，然后看看和前面那个数相同不相同，如果不相同，那么就将这个数加到sum上，然后将cnt=1,如果相同的话，就看看cnt是否下余a[i],如果小于的话，就cnt++,否则的话，就将a[i]加到sum上，cnt=1!
注意：输入的数为除了本身之外，所以要加1！

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[55];
int main()
{
int T,N;
scanf("%d",&T);
N=T;
while(T--)
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]+=1;
}
printf("Case %d: ",N-T);
sort(a+1,a+n+1);
int sum=a[1],cnt=1;
for(int i=2;i<=n;i++)
{
if(a[i]!=a[i-1])
{
sum+=a[i];
cnt=1;
}
else
{
if(cnt<a[i])
cnt++;
else
{
sum+=a[i];
cnt=1;
}
}
}
printf("%d\n",sum);
}
return 0;
}