[hdu5400 Arithmetic Sequence]预处理，容斥

题意：http://acm.hdu.edu.cn/showproblem.php?pid=5400

思路：预处理出每个点向左和向右的最远边界，从左向右枚举中间点，把区间答案加到总答案里面。由与可能与前面的区间重叠，需要减去重复的答案，由于左边界非降，所以重叠的区间长度很容易得到。

#pragma comment(linker, "/STACK:10240000")

#include <map>

#include <set>

#include <cmath>

#include <ctime>

#include <deque>

#include <queue>

#include <stack>

#include <vector>

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

#define X                   first

#define Y                   second

#define pb                  push_back

#define mp                  make_pair

#define all(a)              (a).begin(), (a).end()

#define fillchar(a, x)      memset(a, x, sizeof(a))

#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;

typedef pair<int, int> pii;

typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE

void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}

void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>

void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;

while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>

void print(const T t){cout<<t<<endl;}template<typename F,typename...R>

void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>

void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}

//#endif

template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}

template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);

const int INF = 1e9 + ;

const double EPS = 1e-12;

/* -------------------------------------------------------------------------------- */

const int maxn = 1e5 + ;

int a[maxn], R[maxn], L[maxn];

int main() {

#ifndef ONLINE_JUDGE

    freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", stdout);

#endif // ONLINE_JUDGE

    int n, d1, d2;

    while (cin >> n >> d1 >> d2) {

        for (int i = ; i <= n; i ++) {

            scanf("%d", a + i);

        }

        for (int i = n; i >= ; i --) {

            a[i] = a[i] - a[i - ];

        }

        a[n + ] = INF * ;

        R[n + ] = INF;

        for (int i = ; i <= n; i ++) {

            L[i] = max(, a[i] == d1? L[i - ] : i);

        }

        for (int i = n; i; i --) {

            R[i] = min(n, a[i + ] == d2? R[i + ] : i);

        }

        int lastl = , lastr =;

        ll ans = ;

        for (int i = ; i <= n; i ++) {

            ll c = R[i] - L[i] + ;

            ans += c * (c + ) / ;

            ll cc = min(lastr, R[i]) - L[i] + ;

            if (cc > ) ans -= cc * (cc + ) / ;

            lastl = L[i];

            umax(lastr, R[i]);

        }

        cout << ans << endl;

    }

    return ;

}