CodeForces 1132B Discounts

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 Love2024-07-165.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 Curling2024-08-01

19.CodeForces 1132B Discounts2024-08-01

20.CodeForces 1619D New Year's Problem2024-08-0121.Codeforces Round 964 (Div. 4)2024-08-07

收起

题目链接：CodeForces 1132B【Discounts】

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思路

       因为使用coupons购买q[i]块巧克力，不需要付最便宜的那块巧克力的钱，所以为了使得优惠最大化，所以可以在使用优惠券的时候购买最贵的p[i]块巧克力，所以计算所有巧克力价格高之和和排序后很快能得到答案。

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代码

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define ll long long
const int N = 3e5 + 10;
 
ll a[N], q[N], sum, n, m;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> q[i];
  }
  sort(a + 1, a + 1 + n);
 
  for (int i = 1; i <= m; i++) {
    cout << sum - a[n - q[i] + 1] << endl;
  }
}
 
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
 
  return 0;
}