Description

Solution

首先發現答案就是將Ti$T_i$進行輪換之後的max{Ti−i}+n−1$max\{T_i-i\}+n-1$的最小值（考慮在起點一次性等待很長時間後，直接不停留地一路走過去）
考慮到環的條件非常麻煩，所以考慮將序列倍長，設Pi=Ti−i$P_i=T_i-i$，那麼答案就變成了：mini=1n{maxj=1n{Pi+j−1}+i}+n−1${\displaystyle \underset{i=1}{\overset{n}{min}}\{\underset{j=1}{\overset{n}{max}}\{P_{i+j-1}\}+i\}+n-1}$
發現如果j$j$列舉到2n−i+1$2n-i+1$不會影響答案。
用線段樹維護區間內的maxP$maxP$和minmidi=l{i+maxriPi}$\underset{i=l}{\overset{mid}{min}}\{i+$

Code

/************************************************
 * Au: Hany01
 * Date: Apr 23rd, 2018
 * Prob: [BZOJ5286][HNOI2018] 轉盤
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef
 
 long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
 

#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 100005;

int mn[maxn << 4], mx[maxn << 4], n;

#define mid ((l + r) >> 1)
#define lc (t << 1)
#define rc (lc | 1)

int query(int t, int l, int r, int val)
{
    if (l == r) return l + max(val, mx[t]);
    if (val >= mx[rc]) return min(query(lc, l, mid, val), mid + 1 + val);
    return min(mn[t], query(rc, mid + 1, r, val));
}

void update(int t, int l, int r, int x, int dt)
{
    if (l == r) { mx[t] = dt; return ; }
    if (x <= mid) update(lc, l, mid, x, dt);
    else update(rc, mid + 1, r, x, dt);
    mx[t] = max(mx[lc], mx[rc]);
    if (l <= n) mn[t] = query(lc, l, mid, mx[rc]);
}

int main()
{
#ifdef hany01
    File("bzoj5286");
#endif

    static int m, p, x, y, las;

    n = read(), m = read(), p = read();
    For(i, 1, n)
        x = read(), update(1, 1, n << 1, i, x - i), update(1, 1, n << 1, i + n, x - i - n);
    printf("%d\n", las = (mn[1] + n - 1));

    while (m --)
        x = read() ^ (las * p), y = read() ^ (las * p),
        update(1, 1, n << 1, x, y - x), update(1, 1, n << 1, n + x, y - x - n),
        printf("%d\n", las = (mn[1] + n - 1));

    return 0;
}
//牆角數枝梅，凌寒獨自開。
//    -- 王安石《梅》