CF1244E

题目

E. Minimizing Difference

给出n个正整数，最多执行k次操作。每次操作可以选择一个数加一或减一。问操作后最小值和最大值的最小差值是多少。

求解

贪心。

每次肯定选择最小值加一或者最大值减一。选哪一个取决于最小值的数量少还是最大值的数量少。假设最大值有x个，那么我们得将这x个最大值都减1才能使最终贡献减一，最小值同理。因此不断贪心合并即可。

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
void solve()
{
    int n;
    long long k;
    scanf("%d %lld", &n, &k);
    map<int, int> mp;
    for (int i = 1; i <= n; i++)
    {
        int temp;
        scanf("%d", &temp);
        mp[temp]++;
    }

    while (1)
    {
        auto it1 = mp.begin();
        int mn = it1->first;
        int mn_num = it1->second;

        auto it2 = mp.rbegin();
        int mx = it2->first;
        int mx_num = it2->second;

        if (mn == mx)
        {
            printf("0\n");
            return;
        }
        if (k < min(mn_num, mx_num))
            break;

        if (mn_num < mx_num)
        {
            it1++;
            int val = it1->first;
            int times = min(1ll * val - mn, k / mn_num);
            k -= 1ll * times * mn_num;
            mp[mn + times] += mp[mn];
            mp.erase(mn);
        }
        else
        {
            it2++;
            int val = it2->first;
            int times = min(1ll * mx - val, k / mx_num);
            k -= 1ll * times * mx_num;
            mp[mx - times] += mp[mx];
            mp.erase(mx);
        }
    }
    int mn = mp.begin()->first;
    int mx = mp.rbegin()->first;
    printf("%d\n", mx - mn);
}

int main()
{
    solve();
    return 0;
}