CodeForces 1992B Angry Monk

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-163.CodeForces 1992C Gorilla and Permutation2024-07-164.CodeForces 1992D Test of Love2024-07-16

5.CodeForces 1992B Angry Monk2024-07-16

6.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 1992B【Angry Monk】

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思路

       找出最大的元素，除了最大的元素的所有元素全部分成1，所以操作次数为sum(数组) - max(数组) - (数组元素个数 - 1)，然后全部合并到最大的元素上，操作次数为sum(数组) - max(数组)，所以最最后得到sum(数组) * 2 - max(数组) * 2 - (数组元素个数 - 1)。

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代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 10;
 
ll a[N];
void solve() {
  ll n, m, maxn = 0, sum = 0;
  cin >> m >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
    maxn = max(maxn, a[i]);
  }
  cout << (2 * sum - 2 * maxn - n + 1) << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
 
  return 0;
}