CodeForces 1992C Gorilla and Permutation

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-16

3.CodeForces 1992C Gorilla and Permutation2024-07-16

4.CodeForces 1992D Test of Love2024-07-165.CodeForces 1992B Angry Monk2024-07-166.CodeForces 1992A Only Pluses2024-07-167.CodeForces 1983B Corner Twist2024-07-168.CodeForces 1983A Array Divisibility2024-07-169.CodeForces 1983C Have Your Cake and Eat It Too2024-07-1610.CodeForces 1883B Chemistry2024-07-2611.CodeForces 1883A Morning2024-07-2612.CodeForces 1883C Raspberries2024-07-2813.CodeForces 1883D In Love2024-07-2814.CodeForces 1883E Look Back2024-07-2815.CodeForces 1883F You Are So Beautiful2024-07-2816.CodeForces 1883G1 Dances (Easy version)2024-07-2817.CodeForces 908B New Year and Buggy Bot2024-08-0118.CodeForces 908C New Year and Curling2024-08-0119.CodeForces 1132B Discounts2024-08-0120.CodeForces 1619D New Year's Problem2024-08-0121.Codeforces Round 964 (Div. 4)2024-08-07

收起

题目链接：CodeForces 1992C【Gorilla and Permutation】

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思路

       根据题意只需要使得f(x)尽可能大，g(x)尽可能小，所以需要将大于等于n的数组排在序列的前端，且按由大到小的顺序依次排列，将小于等于m的数字排在序列的后端，且按从小到大的顺序依次排列。

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代码

#include <functional>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define ll long long
const int N = 500 + 10;
 
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = n; i > m; i--) {
      cout << i << " ";
    }
    for (int i = 1; i <= m; i++) {
      cout << i << " ";
    }
    cout << endl;
  }
  return 0;
}