Be Geeks!

Subject:
https://ac.nowcoder.com/acm/contest/7817/B

Give you $n$个数，设$G(i,j)=gcd(a_i,a_{i+1}...a_j)$，$M(i,j)=max(a_i,a_{i+1}...a_j)$，计算
$$\sum _{1\le i\le j\le n}G(i,j)\cdot M(i,j)$$

Idea:
There are two problems to be solved here, one is the interval $gcd$ , one is the sum of interval maximum values.

• First for $For\; gcd$ , it can be$nlo{g}^m$ , preprocessed,$m$是$a_i$The maximum value. Assuming that the right endpoint is determined, the more the left endpoint goes to the left $gcd\; is$ not strictly decreasing, if it decreases every time it is divided by at least$2$ , sogcd gcdThe number of$g$$c$$d$ is$log$ level. So the interval can then pass$dp$ preprocessed eachgcd gcdThe leftmost endpoint of$g$$c$$d$ .
• Interval $After\; gcd\; is$ preprocessed, enumerate the right endpoint$i$ , and then enumerate eachgcd gcdthe leftmost end pointjj of$g$$c$$d$$j$ , assuming the last left endpoint is$k$ , what we want to calculate is
  $$\sum _{l=j}^{k-1}G(l,i)\cdot M(l,i)$$
  $G(l,i)$ are all the same. Now we need to calculate the sum of the maximum value. A prefix sum can be preprocessed. This prefix sum is based on$i$ is the longest increasing sequence from the right end to the left, and the contribution of each position is the distance between the value at this position and the first value greater than it on the left. Then mess around with calculations.

#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;
const int N=200009;
struct node {
    
    
    int l;
    ll val;
    node(int x,ll y):l(x),val(y) {
    
    }
};
int n,dp[N];
ll a[N],sum[N],ans=0;
vector<node>w[N];

namespace ST {
    
    
int a[N],n,dp[N][25],index[N][25],mm[N];//mm:logn/log2
void ST() {
    
    
    mm[0]=-1;
    for(int i=1; i<=n; ++i) {
    
    
        dp[i][0]=a[i];
        index[i][0]=i;

        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
    }
    for(int j=1; j<=mm[n]; ++j)
        for(int i=1; (i+(1<<j)-1)<=n; ++i) {
    
    
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
            if(dp[i][j-1]<=dp[i+(1<<(j-1))][j-1]) {
    
    
                index[i][j]=index[i+(1<<(j-1))][j-1];
            } else {
    
    
                index[i][j]=index[i][j-1];
            }
        }

}
int query(int x,int y) {
    
    
    int t=mm[y-x+1];
    //return max(dp[x][t],dp[y-(1<<t)+1][t]);
    if(dp[x][t]<=dp[y-(1<<t)+1][t])
        return index[y-(1<<t)+1][t];
    else return index[x][t];
}
}



int main() {
    
    
    scanf("%d",&n);
    ST::n=n;
    for(int i=1; i<=n; i++)
        scanf("%lld",&a[i]),ST::a[i]=a[i];
    ST::ST();
    a[0]=mod;
    for(int i=1; i<=n; i++) {
    
    
        int tmp=i-1;
        while(a[tmp]<=a[i])
            tmp=dp[tmp];
        dp[i]=tmp;
    }
    sum[0]=0;
    for(int i=1; i<=n; i++)
        sum[i]=(sum[dp[i]]+(i-dp[i])*a[i]%mod)%mod;
    for(int i=1; i<=n; i++) {
    
    
        int si1=1;
        w[i].push_back(node(i,a[i]));
        int si=w[i-1].size();
        for(int j=0; j<si; j++) {
    
    
            ll tmp=__gcd(a[i],w[i-1][j].val);
            if(tmp==w[i][si1-1].val)
                w[i][si1-1].l=w[i-1][j].l;
            else
                w[i].push_back(node(w[i-1][j].l,tmp)),si1++;
        }
    }
    /*for(int i=1; i<=n; i++)
        for(int j=0; j<w[i].size(); j++)
            printf("%d %d %lld\n",i,w[i][j].l,w[i][j].val);*/
    for(int i=1; i<=n; i++) {
    
    
        int si=w[i].size();
        ll tmp=0;
        for(int j=0; j<si; j++) {
    
    
            int locate=ST::query(w[i][j].l,i);
            ll ans1;
            ll Sum=(sum[i]-sum[locate]+mod+a[locate]*(locate-w[i][j].l+1)%mod)%mod;
            ans1=(Sum*w[i][j].val%mod-tmp*w[i][j].val%mod+mod)%mod;
            ans=(ans+ans1)%mod;
            tmp=Sum;
            //printf("%d %d %d %lld %lld\n",i,w[i][j].l,locate,w[i][j].val,ans1);
        }
    }
    printf("%lld",ans);
    return 0;
}