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 every1 ≤ 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 aand b you are to find sum of products ca·cb. Will you help her?
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.
Print the only integer — summary price of passing each of the roads in XXX.
4 1
2 3 1 2
3
17
5 2
3 5 2 2 4
1 4
71
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.
有一点意思的模拟
n个城市形成一个环，一共n条边。此外这里头又有m个'vip城市'保证到其他城市都有边，一条连接i和j的边长度就是v[i]*v[j]，问所有边的总长
其实还是sb题嘛
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set> 10 #include<map> 11 #include<ctime> 12 #define LL long long 13 #define inf 0x7ffffff 14 #define pa pair<int,int> 15 #define pi 3.1415926535897932384626433832795028841971 16 using namespace std; 17 inline LL read() 18 { 19 LL x=0,f=1;char ch=getchar(); 20 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 21 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 22 return x*f; 23 } 24 inline void write(LL a) 25 { 26 if (a<0){printf("-");a=-a;} 27 if (a>=10)write(a/10); 28 putchar(a%10+'0'); 29 } 30 inline void writeln(LL a){write(a);printf(" ");} 31 int n,m; 32 LL a[100010],b[100010]; 33 LL ans,v; 34 bool mrk[100010]; 35 int main() 36 { 37 n=read();m=read(); 38 for (int i=1;i<=n;i++)a[i]=read(),v+=a[i]; 39 for (int i=1;i<=m;i++)b[i]=read(),mrk[b[i]]=1; 40 for (int i=1;i<=m;i++) 41 { 42 ans+=a[b[i]]*(v-a[b[i]]); 43 v-=a[b[i]]; 44 } 45 for(int i=1;i<=n;i++) 46 { 47 int t=i+1;if (t>n)t=1; 48 if (!mrk[t]&&!mrk[i])ans+=a[i]*a[t]; 49 } 50 printf("%lld ",ans); 51 }