POI #Year2012 #二分
考慮二分答案,然後如果 \(|a_i-a_{i-1}|>mid\) ,那麼一定要提前操作掉,先把這種情況搞掉
然後考慮列舉一個位置變成 \(0\) ,在上面的操作後,可以保證 \(|a_i-a_{i-1}|\leq mid\) ,那麼這時還需要操
作的區間 \([l,r]\) ,\(l,r\) 都隨著 \(i\) 的增加而增加,維護這兩個端點直接算答案
// Author: xiaruize
const int N = 1e6 + 10;
int n, m;
int a[N];
int s[N], sum[N];
int check(int x)
{
memcpy(s, a, sizeof(s));
int cnt = 0;
rep(i, 2, n) s[i] = min(s[i - 1] + x, s[i]);
per(i, n - 1, 1) s[i] = min(s[i], s[i + 1] + x);
rep(i, 1, n)
{
cnt += a[i] - s[i];
sum[i] = sum[i - 1] + s[i];
}
for (int i = 1, l = 1, r = 1; i <= n; i++)
{
while (r < n && (r - i + 1) * x < s[r + 1])
r++;
while (l < i && (i - l) * x >= s[l])
l++;
if (cnt + sum[r] - sum[l - 1] - x * ((i - l) * (i - l + 1) / 2 + (r - i) * (r - i + 1) / 2) <= m)
return i;
}
return -1;
}
void solve()
{
cin >> n >> m;
rep(i, 1, n) cin >> a[i];
int l = 0, r = 1e9;
pii res = {-1, -1};
while (l < r)
{
int mid = l + r >> 1;
int v = check(mid);
if (v == -1)
l = mid + 1;
else
{
r = mid;
res = {mid, v};
}
}
cout << res.second << ' ' << res.first << endl;
}
#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
#ifndef ONLINE_JUDGE
cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
return 0;
}