HDU 6917 Shorten the array

题目链接

Problem Description

Alice has an array a. The array has N positive numbers. She thinks the array is too long, so she wants to shorten the array. She decides to shorten the array via the following operation: every time she will choose an index i(1≤i<n)i (1≤i<n)i(1≤i<n) which ai>0ai>0ai>0 and ai+1>0a_i+1>0ai+1>0. She will deleteaia_iai and ai+1a_i+1ai+1 and use (ai(a_i(ai modmodmod ai+1)ai+1)ai+1) or (ai+1(ai+1(ai+1 modmodmod ai)a_i)ai) to replace them.

For example, for array [3,2,4,5][3,2,4,5][3,2,4,5], if Alice choose i=2i=2i=2, she can change the array to [3,2,5][3,2,5][3,2,5] or [3,0,5][3,0,5][3,0,5]. Alice wants you to tell her the shortest possible length of the array after several options.

Input

The first line contains a single integer T(1≤T≤10)T (1≤T≤10)T(1≤T≤10), indicating the number of test cases.

For each test cases:

The first line contains one integer N(2≤N≤106)N (2≤N≤10^6)N(2≤N≤106), indicating the size of the array.

The next line containsNNN integers ai(ai≤109)a_i (a_i≤10^9)ai(ai≤109), representing the array.

Output

Output TTT lines.

The i−thi-thi−th line contains a single integer, representing the answer of i−thi-thi−th test case.

Sample Input

2
4
1 1 1 1
4
2 3 4 5

Sample Output

2
1

Hint

For the first sample, Alice first choose i=1i=1i=1 to change the array to [0,1,1][0, 1, 1][0,1,1], then choose i=2i=2i=2 to change the array to [0,0][0, 0][0,0], which is the best result she can reach.

解题思路：

取模运算有如下性质：
a%b=a(a<b)a\%b=a \ \ (a<b) a%b=a (a<b)
Case1:对于样例2：23452\ 3\ 4\ 52 3 4 5

我们可以用最小的数222对其余数进行取模运算，即2%3,2%4,2%52\%3,2\%4,2\%52%3,2%4,2%5，最后数组元素只剩2，长度为1。

Case2:题目要求做运算的数必须大于0。对于样例1：11111\ 1\ 1\ 11 1 1 1
得到最终结果为000 \ 00 0，数组长度为2

可知，当数组中只存在相同的数时，数组长度为iii与i+1i+1i+1将得到同样的结果，都是(i+1)/2(i+1)/2(i+1)/2。

Case3:对于样例2223452\ 2\ 2\ 3\ 4\ 52 2 2 3 4 5
按如上做法将会得到答案为2(02)2\ (0\ 2)2 (0 2)。但若进行运算时可以得到比最小数(min=2)更小的数，如3%2=1，将此更小的数保存下来（而不是保存2%3=2），此时最小数将更新。于是如此处理后的数组将变为2212\ 2\ 12 2 1
最后可以将其他所有的数吞并，最终答案一定为1。

注意：第二次更新的最小数xxx需满足 (x>0&&x<min)(x>0 \&\&x<min)(x>0&&x<min)，并且只能转换一个更小的数来达到目标。

AC代码：

#include<iostream>
#define int long long
using namespace std;
int a[1000005];
int t,n;
signed main()
{// ATTENTION %LLD FOR LONG LONG !!!scanf("%lld",&t);while(t--){int minn=1000000005;scanf("%lld",&n);for(int i=0;i<n;i++){scanf("%lld",&a[i]);if(minn>a[i])   minn=a[i];}int s=0,f=0;for(int i=0;i<n;i++){if((a[i]%minn>0&&a[i]%minn<minn))//判断运算后是否能生成更小的数//此处只需判断与minn的余数即可{f=1;break;}if(minn==a[i])  s++;}if(f==1)    printf("1\n");else printf("%lld\n",(s+1)/2);}return 0;
}