B. Mishka and trip 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.
題目連接配接：http://codeforces.com/contest/703/problem/B
題意：n個城市，每個城市之間最多有一條道路直接相連。 1 — 2 — ... — n — 1之間有一條道路。其中k個城市是首都，首都與其他城市之間都有道路直接相連。輸出這個城市所有道路的權值。
思路：n個城市的權值總和為sum，每個首都與其他城市之間的道路權值為c[i]*(sum-c[i-1]-c[i+1]-cou)。其中cou為c[i]之前出現過的首都的權值總和。考慮c[i-1]與c[i+1]在i=1和i=n的情況和cou裡面包括了c[i-1]和c[i+1]的情況。
代碼：
#include<bits/stdc++.h>
using namespace std;
__int64 c[100100];
int d[100100];
int sign[100100];
int main()
{
int i,n,k;
scanf("%d%d",&n,&k);
__int64 ans=0,sum=0;
for(i=1; i<=n; i++)
{
scanf("%I64d",&c[i]);
if(i>1) ans+=(c[i-1]*c[i]);
sum+=c[i];
}
ans+=(c[n]*c[1]);
__int64 cou=0;
memset(sign,0,sizeof(sign));
for(i=0; i<k; i++)
{
scanf("%d",&d[i]);
__int64 flag=0;
if(d[i]>1)
{
if(sign[d[i]-1]==0)
flag+=c[d[i]-1];
}
else
{
if(sign[n]==0)
flag+=c[n];
}
if(d[i]<n)
{
if(sign[d[i]+1]==0)
flag+=c[d[i]+1];
}
else
{
if(sign[1]==0)
flag+=c[1];
}
//cout<<c[d[i]]<<" "<<flag<<" "<<cou<<endl;
ans+=c[d[i]]*(sum-flag-cou-c[d[i]]);
cou+=c[d[i]];
sign[d[i]]=1;
}
printf("%I64d\n",ans);
return 0;
} 模拟
轉載于:https://www.cnblogs.com/GeekZRF/p/5747327.html