CodeForces703B Mishka and trip 數學

Megumin發表於2016-08-06

Little Mishka is a great traveller and she visited many countries. After thinking about where to travel this time, she chose XXX — beautiful, but little-known northern country.

Here are some interesting facts about XXX:

  1. XXX consists of n cities, k of whose (just imagine!) are capital cities.
  2. All of cities in the country are beautiful, but each is beautiful in its own way. Beauty value of i-th city equals to ci.
  3. All the cities are consecutively connected by the roads, including 1-st and n-th city, forming a cyclic route 1 — 2 — ... — n — 1. Formally, for every 1 ≤ i < n there is a road between i-th and i + 1-th city, and another one between 1-st and n-th city.
  4. Each capital city is connected with each other city directly by the roads. Formally, if city x is a capital city, then for every 1 ≤ i ≤ n,  i ≠ x, there is a road between cities x and i.
  5. There is at most one road between any two cities.
  6. Price of passing a road directly depends on beauty values of cities it connects. Thus if there is a road between cities i and j, price of passing it equals ci·cj.

Mishka started to gather her things for a trip, but didn't still decide which route to follow and thus she asked you to help her determine summary price of passing each of the roads in XXX. Formally, for every pair of cities a and b (a < b), such that there is a road between a and b you are to find sum of products ca·cb. Will you help her?

Input

The first line of the input contains two integers n and k (3 ≤ n ≤ 100 000, 1 ≤ k ≤ n) — the number of cities in XXX and the number of capital cities among them.

The second line of the input contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 10 000) — beauty values of the cities.

The third line of the input contains k distinct integers id1, id2, ..., idk (1 ≤ idi ≤ n) — indices of capital cities. Indices are given in ascending order.

Output

Print the only integer — summary price of passing each of the roads in XXX.

Examples
input
4 1
2 3 1 2
3
output
17
input
5 2
3 5 2 2 4
1 4
output
71
Note

This image describes first sample case:

It is easy to see that summary price is equal to 17.

This image describes second sample case:

It is easy to see that summary price is equal to 71.


求解首都城市到其他城市的價值之和時，用字首和來優化，併為避免重複，每處理一個首都城市就應該在字首和中將其出去。


#include <iostream>
#include <cstdio>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
int c[100005];
int a[100005];
bool v[100005];
int main()
{
    int n,k;
    LL ans,sum;
    while(~scanf("%d%d",&n,&k)){
        ans=0;
        memset(v,false,sizeof(v));
        scanf("%d",&c[1]);
        sum=c[1];
        for(int i=2;i<=n;i++){
            scanf("%d",&c[i]);
             ans+=c[i]*c[i-1];
            sum+=c[i];
        }
        ans+=c[1]*c[n];
        for(int i=1;i<=k;i++){
            scanf("%d",&a[i]);
            v[a[i]]=true;
            int p=0;
            int q=0;
            if(a[i]==1){
                q=n;
            }else if(a[i]==n){
                p=-n;
            }
            sum-=c[a[i]];
            LL temp=sum;
            if(!v[a[i]+1+p]){
                temp-=c[a[i]+1+p];
            }
            if(!v[a[i]-1+q]){
                temp-=c[a[i]-1+q];
            }
            ans+=temp*c[a[i]];
        }
        cout<<ans<<endl;;
    }
    return 0;
}


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