Codeforces Round #685 (Div. 2) A. Subtract or Divide 构造

Ridbit starts with an integer n.

In one move, he can perform one of the following operations:

divide n by one of its proper divisors, or
subtract 1 from n if n is greater than 1.
A proper divisor is a divisor of a number, excluding itself. For example, 1, 2, 4, 5, and 10 are proper divisors of 20, but 20 itself is not.

What is the minimum number of moves Ridbit is required to make to reduce n to 1?

Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.

The only line of each test case contains a single integer n (1≤n≤109).

Output
For each test case, output the minimum number of moves required to reduce n to 1.

Example
inputCopy
6
1
2
3
4
6
9
outputCopy
0
1
2
2
2
3
Note
For the test cases in the example, n may be reduced to 1 using the following operations in sequence

1
2→1
3→2→1
4→2→1
6→2→1
9→3→2→1
前几个特判一下，然后偶数–2--1，奇数先减1到偶数再—2----1

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<stdlib.h>
 
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int maxn=1e6+5;
const int inf=0x3f3f3f3f;

ll m,n,t,ans,res,tmp,a[maxn]; 
int main()
{
	ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
	cin>>t;
	while(t--)
	{
		cin>>n;
		if(n==1)
			cout<<"0"<<endl;
		else if(n==2)
			cout<<"1"<<endl;
		else if(n==3)
			cout<<"2"<<endl;
		else if(n%2==0)
			cout<<"2"<<endl;
		else
			cout<<"3"<<endl;
	}
	return 0;
}

标签：Divide,int,proper,number,Codeforces,test,685,integer,include
来源： https://blog.csdn.net/qq_45891413/article/details/109945729