time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Given an integer n𝑛, find an integer x𝑥 such that:
- 2≤x≤n2≤𝑥≤𝑛.
- The sum of multiples of x𝑥 that are less than or equal to n𝑛 is maximized. Formally, x+2x+3x+⋯+kx𝑥+2𝑥+3𝑥+⋯+𝑘𝑥 where kx≤n𝑘𝑥≤𝑛 is maximized over all possible values of x𝑥.
Input
The first line contains t𝑡 (1≤t≤1001≤𝑡≤100) — the number of test cases.
Each test case contains a single integer n𝑛 (2≤n≤1002≤𝑛≤100).
Output
For each test case, output an integer, the optimal value of x𝑥. It can be shown there is only one unique answer.
Example
input
Copy
2
3
15
output
Copy
3 2
Note
For n=3𝑛=3, the possible values of x𝑥 are 22 and 33. The sum of all multiples of 22 less than or equal to n𝑛 is just 22, and the sum of all multiples of 33 less than or equal to n𝑛 is 33. Therefore, 33 is the optimal value of x𝑥.
For n=15𝑛=15, the optimal value of x𝑥 is 22. The sum of all multiples of 22 less than or equal to n𝑛 is 2+4+6+8+10+12+14=562+4+6+8+10+12+14=56, which can be proven to be the maximal over all other possible values of x𝑥.
解题说明：此题是一道数学题，为了保证x的倍数之和尽可能大，同时kx小于n，则x应该尽可能小，求和的值为(1+k)/2 * kx 而kx小于n ，就要保证k值越大越好，那么很显然x越小k值容易越大。x最小 为2。此时要区分n为3的情况，此时x为3为最优解。
#include<stdio.h>int main()
{int t;scanf("%d", &t);while (t--){int n;scanf("%d", &n);if (n == 3){printf("3\n");}else{printf("2\n");}}return 0;
}
