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：

 1 class Solution {
 2 public:
 3     int totalStrength(vector<int>& a) {
 4         int n=a.size();
 5         vector<int>left,right;
 6         stack<int>st;
 7         st.push(-1);
 8         for(int i=0;i<n;i++){
 9             while(st.top()>=0 and a[i]<=a[st.top()]){
10                 st.pop();
11             }
12             left.push_back(st.top());
13             st.push(i);
14         }
15         while(!st.empty())st.pop();
16         st.push(n);
17         for(int i=n-1;i>=0;i--){
18             while(st.top()<n and a[i]<a[st.top()]){
19                 st.pop();
20             }
21             right.push_back(st.top());
22             st.push(i);
23         }
24         reverse(right.begin(),right.end());
25         int mod=1e9+7;
26         int res=0;
27         vector<long long>pre(n+1,0);
28         for(int i=0;i<n;i++){
29             pre[i+1]=(pre[i]+a[i])%mod;
30         }
31         vector<long long>prepre(n+2,0);
32         for(int i=1;i<=n;i++){
33             prepre[i+1]=(prepre[i]+pre[i])%mod;
34         }
35         for(int i=0;i<n;i++){
36             int l=left[i]+1;
37             int r=right[i]-1;
38             int tmp=(prepre[r+2]-prepre[i+1]+mod)%mod*(i-l+1)%mod;
39             int tmp2=(prepre[i+1]-prepre[l]+mod)%mod*(r-i+1)%mod;
40             res=(res+(long long)a[i]*(tmp-tmp2+mod)%mod)%mod;
41         }
42         return res;
43     }
44 };

标签：pre,right,totalStrength,leetcode2281,巫师,st,int,管辖,力量
来源： https://www.cnblogs.com/wangyiming/p/16320937.html