牛客小白月赛27 I -名作之壁 （单调队列，双指针）
题面：
思路：
我们很容易知道，当确定一个区间的左端点(mathit l) 后，区间最小值和最大值是关于区间右端点$mathit r $ 的一个单峰函数。
那么对于一个左端点(mathit l)，我们只需要找到一个最小的$mathit r $ 使其最大值和最小值之差大于$mathit k $ ，那么([r,n])中每一个值作右端点都是合法的。直接对答案加上贡献，然后将左端点右推一位，继续找到其对应的最小右端点即可。
那么如何快速判断区间中最大值和最小值之差呢？
我们可以用两个单调队列，分别维护数组中单调递减和单调递增的数值的下标，
然后在枚举区间右端点的过程中维护单调性，在左端点右移的时候维护队列中下标要在当前区间中即可。
时间复杂度为：(O(n))
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n, k;
int a[10000010];
deque<int> mi, ma;
const int mod = 1e9;
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
n = readint();
k = readint();
a[0] = readint();
int b = readint();
int c = readint();
repd(i, 1, n)
{
a[i] = (1ll * a[i - 1] * b + c) % mod;
}
ll ans = 0ll;
int l = 1;
repd(i, 1, n)
{
while (!mi.empty() && a[mi.back()] > a[i])
{
mi.pop_back();
}
while (!ma.empty() && a[ma.back()] < a[i])
{
ma.pop_back();
}
mi.push_back(i);
ma.push_back(i);
while (!mi.empty() && !ma.empty() && a[ma.front()] - a[mi.front()] > k)
{
ans += n - i + 1;
l++;
while (ma.front() < l )
{
ma.pop_front();
}
while (mi.front() < l )
{
mi.pop_front();
}
}
}
printf("%lld
", ans );
return 0;
}