Codefores 1151E Number of Components (计数)

大意:给定n元素序列$a$, $1le a_i le n$, 定义函数$f(l,r)$表示范围在$[l,r]$以内的数构成的连通块个数, 求$sumlimits_{i=1}^{n}sumlimits_{j=i}^{n}f(i,j)$

对于序列$a$中一个区间$[l,r]$, 假设最小值$mi$, 最大值$ma$, 它要想构成一个连通块的充要条件是$a[l-1],a[r+1]$不在$[mi,ma]$范围内, 可以得到贡献为$mi(n-ma+1)$. 但是显然不能暴力枚举所有区间, 我们可以枚举合法区间的右端点来计算.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+10;
int n, m, k, t;
int a[N];


int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i);
	ll ans = (ll)a[n]*(n-a[n]+1);
	REP(i,1,n-1) {
		if (a[i]<a[i+1]) ans+=(ll)a[i]*(a[i+1]-a[i]);
		else ans+=(ll)(a[i]-a[i+1])*(n-a[i]+1);
	}
	printf("%lld
", ans);
}