牛可乐发红包脱单ACM赛$

B 小a的旅行计划
oeis
http://oeis.org/search?q=0%2C3%2C30+&language=english&go=Search

int main(void)
{
    LL n;
    cin>>n;
    n++;
    LL ans = (qpow(4,n)-4*qpow(3,n)+6*qpow(2,n)-4)*qpow(8,mod-2);
    ans = (ans%mod+mod)%mod;
    cout<<ans<<endl;
   return 0;
}

C 区区区间间间

$v_{l, r} = m a x (a_{i} - a_{j}) (l < = i, j < = r)$
$<munderover> \sum i n </munderover> <munderover> \sum j + 1 n </munderover> v_{i, j}$
对于每一个a[i] 左边有连续的x个比它小，右边有连续的y个比它大
那么a[i] 作为这个区间的最大值，被统计的次数就是(x+1)(y+1)-1

通过单调队列可以求出x和y



#include <bits/stdc++.h>
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define Pb push_back
#define FI first
#define SE second
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define IOS ios::sync_with_stdio(false)
#define DEBUG cout<<endl<<"DEBUG"<<endl; 
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int    prime = 999983;
const int    INF = 0x7FFFFFFF;
const LL     INFF =0x7FFFFFFFFFFFFFFF;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-6;
const LL     mod = 1e9 + 7;
LL qpow(LL a,LL b){LL s=1;while(b>0){if(b&1)s=s*a%mod;a=a*a%mod;b>>=1;}return s;}
LL gcd(LL a,LL b) {return b?gcd(b,a%b):a;}
int dr[2][4] = {1,-1,0,0,0,0,-1,1};
typedef pair<int,int> P;
const int maxn = 1e5+100;
int a[maxn];
int s[maxn];// 单调栈
// 第一遍求在这个区间里面最大
int pre[maxn];
int nxt[maxn];
int main(void)
{
    int T,n;
    cin>>T;
    while(T--){
        scanf("%d",&n);
        for(int i = 1;i <= n; ++i){
            scanf("%d",&a[i]);
        }
        int t = 0;
        for(int i = 1;i <= n; ++i){
            pre[i] = nxt[i] = 0;
            while(t > 0&&a[i] > a[s[t]]) nxt[s[t]] = i,t--;
            pre[i] = s[t];
            s[++t] = i;
            // cout<<pre[i]<<" ";
        }
        while(t > 0)
            nxt[s[t]] = n+1,t--;
        LL ans = 0;
        for(int i = 1;i <= n; ++i){
            ans += 1ll*a[i]*(nxt[i]-i)*(i-pre[i]);
        }
        t = 0;
        for(int i = 1;i <= n; ++i){
            pre[i] = nxt[i] = 0;
            while(t > 0&&a[i] < a[s[t]]) nxt[s[t]] = i,t--;
            pre[i] = s[t];
            s[++t] = i;
        }
         while(t > 0)
            nxt[s[t]] = n+1,t--;
        for(int i = 1;i <= n; ++i){
            ans -= 1ll*a[i]*(nxt[i]-i)*(i-pre[i]);
        }
        printf("%lld\n",ans);
    }

    return 0;
}