leetcode2281 巫师的总力量和

思路：

单调栈计算出每个数字的“管辖范围”，再对每个“管辖范围”单独计算并求和。

实现：

class Solution {
public:
    int totalStrength(vector<int>& a) {
        int n=a.size();
        vector<int>left,right;
        stack<int>st;
        st.push(-1);
        for(int i=0;i<n;i++){
            while(st.top()>=0 and a[i]<=a[st.top()]){
                st.pop();
            }
            left.push_back(st.top());
            st.push(i);
        }
        while(!st.empty())st.pop();
        st.push(n);
        for(int i=n-1;i>=0;i--){
            while(st.top()<n and a[i]<a[st.top()]){
                st.pop();
            }
            right.push_back(st.top());
            st.push(i);
        }
        reverse(right.begin(),right.end());
        int mod=1e9+7;
        int res=0;
        vector<long long>pre(n+1,0);
        for(int i=0;i<n;i++){
            pre[i+1]=(pre[i]+a[i])%mod;
        }
        vector<long long>prepre(n+2,0);
        for(int i=1;i<=n;i++){
            prepre[i+1]=(prepre[i]+pre[i])%mod;
        }
        for(int i=0;i<n;i++){
            int l=left[i]+1;
            int r=right[i]-1;
            int tmp=(prepre[r+2]-prepre[i+1]+mod)%mod*(i-l+1)%mod;
            int tmp2=(prepre[i+1]-prepre[l]+mod)%mod*(r-i+1)%mod;
            res=(res+(long long)a[i]*(tmp-tmp2+mod)%mod)%mod;
        }
        return res;
    }
};