Codeforces Round #365 (Div. 2) -- B. Mishka and trip （找规律枚举）

[align=center]B. Mishka and trip[/align]

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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:

XXX consists of n cities,
k of whose (just imagine!) are capital cities.
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.
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.

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.

There is at most one road between any two cities.

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.

大体题意:

有n个城市，普通城市会和下一个城市有一条连线，省会城市 会与其他所有城市都有一条边，边的权值是两个城市权值的乘积，求所有边的权值之和！

思路：

看第二个样例吧，省会城市有1和4，那么先算1相连的，计算是：

a1*a2 + a1*a3 + a1*a4+a1*a5 = a1*(a2+a3+a4+a5)所以可以先预处理一个权值之和，第一个城市相连边权值之和就是 (sum - a[1]) * a[1];

在算城市4，计算要减去重复的，计算式是 (sum-a[1]-a[4])*a[4]  这就是规律了。

这样算完后 会发现城市2还没有访问，所以可以在遍历一遍n个城市，求出未访问的城市，单独在计算一遍即可！

详细见代码：

#include<bits/stdc++.h>

using namespace std;
const int maxn = 100000 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef long long ll;
int a[maxn];
int b[maxn];
ll sum[maxn];
ll sum2[maxn];
int vis[maxn];
int main(){
int n,m;
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; ++i)scanf("%d",&a[i]);
for (int i = 1; i <= n; ++i){
sum[i] = sum[i-1] + a[i]*a[i == n ? 1 : (i+1)];
sum2[i] = sum2[i-1] + a[i];
}

for (int i = 1; i <= m; ++i)scanf("%d",&b[i]);
// printf("%d\n",sum
);
ll ans = 0ll;
ll sumcnt = 0ll;
for (int i = 1; i <= m; ++i){
ans += (sum2
- a[b[i]] - sumcnt) * a[b[i]];
sumcnt += a[b[i]];
vis[b[i] ] = 1;
int down;
if (b[i] == 1)down = n;
else down = b[i]-1;
vis[down] = 1;
}
for (int i = 1; i <= n; ++i){
if (vis[i])continue;
ans += a[i]*a[i == n ? 1 : (i+1)];

}

printf("%I64d\n",ans);
return 0;
}