MG loves gold

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 982    Accepted Submission(s): 420

Problem Description

MG is a lucky boy. He is always be able to find gold from underground.

It is known that the gold is a sequence with n elements, which has its own color C.

MG can dig out a continuous area of sequence every time by using one shovel, but he's unwilling to dig the golds of the same color with one shovel.

As a greedy person, he wish to take all the n golds away with least shovel.
The rules also require MG not to dig twice at the same position.

MG thought it very easy and he had himself disdained to take the job. As a bystander, could you please help settle the problem and calculate the answer?

 

Input

The first line is an integer T which indicates the case number.(1<=T<=10)

And as for each case, there are 1 integer n in the first line which indicate gold-number(1<=n<=100000).

Then there are n integers C in the next line, the x-th integer means the x-th gold’s color(|C|<=2000000000).

 

Output

As for each case, you need to output a single line.

there should be one integer in the line which represents the least possible number of shovels after taking away alln golds.

 

Sample Input

251 1 2 3 -151 1 2 2 3

 

Sample Output

23

#include<stdio.h>#include<math.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int main(){    long long T,n,i,sum,j,k;    long long a[100001];    while(scanf("%lld",&T)!=EOF)    {        while(T--)        {            sum=1;            scanf("%lld",&n);            for(i=0;i<n;i++)            {                scanf("%lld",&a[i]);            }            j=0;k=0;            while(1)            {                j++;                if(j>=n)                    break;                for(i=j-1;i>=k;i--)                {                    if(a[j]==a[i])                    {                        sum++;                        k=j;                        break;                    }                }            }            printf("%lld\n",sum);        }    }    return 0;}