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

A. Greatest Convex

You are given an integer $k$. Find the largest integer $x$, where $1\le x<k$, 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 $y!=y\cdot (y-1)!$ for $y\ge 1$ with the base case of $0!=1$. For example, $5!=5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 0!=120$.

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

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

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

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$1\le x<k$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$(2\le k\le {10}^9)$，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$x!+(x-1)!=(x+1)\ast (x-1)!=k\ast (k-2)!$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