
给你一个数组,你最多可以进行k次操作,每次操作可以使一个数+1或者-1,问操作之后数组的极差最小可能是多少
利用map来模拟移动,可以观察到每次应该选择数量少的一组数让他们进行移动是最优的
int main(){
int n;
ll k;
cin >> n >> k;
vector<int> a(n);
map<int,int> ls;
for(int i = 0 ; i < n ; i++)
cin >> a[i],ls[a[i]]++;
while(ls.size() > 1) {
auto l = ls.begin();
auto r = ls.end();
--r;
auto nl = ++l;
--l;
auto nr = --r;
r++;
if(l->se <= r-> se) {
ll cost = l->se * 1ll * (nl->fi - l->fi);
if(k <= cost) break;
k -= cost;
nl->se += l->se;
ls.erase(l);
}
else {
ll cost = r->se * 1ll * (r->fi - nr->fi);
if(k <= cost) break;
k -= cost;
nr->se += r->se;
ls.erase(r);
}
}
int c1 = ls.begin()->se;
int c2 = ls.rbegin()->se;
int mins = k / min(c1 , c2);
cout << max(0 , ls.rbegin()->fi - ls.begin()->fi - mins) << endl;
}
Codeforces Round #592 (Div. 2) E
原文:https://www.cnblogs.com/033000-/p/12310444.html