比较巧妙的地方是将询问映射到二维坐标系里，虽然比较套路但考场上不一定能想到。剩下的就是主席树裸题了。
代码用时：40min。
#include<cstdio>
#include<vector>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std; const int N=;
int n,m,p1,p2,l,r,cnt,top,rt[N],a[N],L[N],R[N],stk[N],v[N*],ls[N*],rs[N*];
ll s[N*];
struct P{ int x,y; ll p; };
vector<P>V[N]; void mdf(int y,int &x,int L,int R,int l,int r,int k){
v[x=++cnt]=v[y]; s[x]=s[y]; ls[x]=ls[y]; rs[x]=rs[y];
s[x]+=1ll*(r-l+)*k;
if (l==L && r==R){ v[x]+=k; return; }
int mid=(L+R)>>;
if (r<=mid) mdf(ls[y],ls[x],L,mid,l,r,k);
else if (l>mid) mdf(rs[y],rs[x],mid+,R,l,r,k);
else mdf(ls[y],ls[x],L,mid,l,mid,k),mdf(rs[y],rs[x],mid+,R,mid+,r,k);
} ll que(int x,int L,int R,int l,int r){
if (!x) return ;
if (l==L && r==R) return s[x];
int mid=(L+R)>>; ll sum=1ll*(r-l+)*v[x];
if (r<=mid) return sum+que(ls[x],L,mid,l,r);
else if (l>mid) return sum+que(rs[x],mid+,R,l,r);
else return sum+que(ls[x],L,mid,l,mid)+que(rs[x],mid+,R,mid+,r);
} int main(){
scanf("%d%d%d%d",&n,&m,&p1,&p2);
rep(i,,n) scanf("%d",&a[i]);
rep(i,,n){
while (top && a[stk[top]]<a[i]) top--;
L[i]=stk[top]; stk[++top]=i;
}
rep(i,,top) stk[i]=; stk[top=]=n+;
for (int i=n; i; i--){
while (top && a[stk[top]]<a[i]) top--;
R[i]=stk[top]; stk[++top]=i;
}
rep(i,,n){
if (i<n) V[i].push_back((P){i+,i+,p1});
if (L[i] && R[i]<=n) V[L[i]].push_back((P){R[i],R[i],p1});
if (L[i] && i<=R[i]-) V[L[i]].push_back((P){i+,R[i]-,p2});
if (R[i]<=n && i>=L[i]+) V[R[i]].push_back((P){L[i]+,i-,p2});
}
rt[]=;
rep(i,,n){
rt[i]=rt[i-];
for (vector<P>::iterator it=V[i].begin(); it!=V[i].end(); it++)
mdf(rt[i],rt[i],,n,it->x,it->y,it->p);
}
rep(i,,m) scanf("%d%d",&l,&r),printf("%lld\n",que(rt[r],,n,l,r)-que(rt[l-],,n,l,r));
return ;
}