A. Greatest Convex【Codeforces Round #842 (Div. 2)】

A. Greatest Convex

You are given an integer \(k\). Find the largest integer \(x\), where \(1≤x<k\), such that \(x!+(x−1)!\)† is a multiple of ‡ \(k\), or determine that no such \(x\) exists.

† \(y!\) denotes the factorial of \(y\), which is defined recursively as \(y!=y⋅(y−1)!\) for \(y≥1\) with the base case of \(0!=1\). For example, \(5!=5⋅4⋅3⋅2⋅1⋅0!=120\).

‡ If \(a\) and \(b\) are integers, then \(a\) is \(a\) multiple of \(b\) if there exists an integer \(c\) such that \(a=b⋅c\). For example, \(10\) is a multiple of \(5\) but \(9\) is not a multiple of \(6\).

Input
The first line contains a single integer \(t\) \((1≤t≤10^4)\) — the number of test cases. The description of test cases follows.

The only line of each test case contains a single integer \(k\) \((2≤k≤10^9)\).

Output
For each test case output a single integer — the largest possible integer \(x\) that satisfies the conditions above.

If no such \(x\) exists, output \(−1\).

Example
input

4
3
6
8
10

output

2
5
7
9

Note
In the first test case, \(2!+1!=2+1=3\), which is a multiple of \(3\).

In the third test case, \(7!+6!=5040+720=5760\), which is a multiple of \(8\).

原题链接

简述题意

给出\(t\)个\(k\)，找出是否存在一个\(x\)满足\(1≤x<k\)且\(x!+(x−1)!\) % \(k==0\)，输出\(x\)的最大值，否则输出\(-1\)

思路

1. 由于\(k\)的范围是\((2≤k≤10^9)\)，因此不能直接枚举求阶乘
2. 观察example的input和output数据的特性我们可以猜测总是存在最大的\(x\)使得\(x = k - 1\)满足条件
3. 当\(x = k - 1\)时\(x! + (x − 1)! = (x + 1) * (x - 1)! = k * (k - 2)!\)总是满足是k的倍数

代码

点击查看代码

#include<iostream>
using namespace std;
int k,t;

int main(){
	cin >> t;
	while(t -- ){
		cin >> k;
		cout << k - 1 << endl;
	}
}

解题历程

1. 错误方向浪费大量时间：多种方式求阶乘，分解质因数，二分搜索
2. Runtime error on pretest 2(218 ms,262100 KB) 【00:19】//阶乘
3. Wrong answer on pretest 1(0 ms,3900 KB) 【00:45】 //分解质因数
4. Compilation error(0 ms,0 KB) 【00:55】 //看出规律，蒙出答案k-1
5. AC(46 ms,0 KB) 【00:57】

经验总结

1. 仔细思考数据范围与关系式的关系
2. 签到题就会有签到题的样子：代码量小，如果代码量大了一个认真分析是否方向错误
3. 可以看排名中的ac时间确定题目的难易
4. 注意对已知关系式的进一步解析
5. Codeforces:†, ‡, §, ¶分别代表1,2,3,4，一般是对题目信息的补充

原因

1. 不熟悉codeforces
2. 手疏