B. Kevin and Permutation codeforces1754B

Problem - B - Codeforces

题目大意：求一个n的排列使得差分的最小值最小

1<=n<=1000

思路：因为n个数每两个相邻的数都要求差，所以差分的最小值就是n/2，那么我们可以输出一个i然后输出i+n/2这样两个数一组输出，如果n为奇数的话，那我们可以从大数开始输出然后最后再单独输出n，比如n=11时，我们的输出就是5,10,4,9,3,8,2,7,1,6,11

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{int t;cin >> t;while (t--){int n;scanf("%d", &n);for (int i = 1; i <= n / 2; i++){//每两个数一组输出，每组之间先大后小printf("%d %d ", n / 2 - (i - 1), n / 2 * 2 - (i - 1));//每组的两个数先小后大}if (n & 1){printf("%d", n);}printf("\n");}return 0;
}

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